This is the best explanation on the internet. he explains why we don;t have coriolis force at equator, why the coriolis force is to your right in NH, and left in SH. also explain clearly that the coriolis force is not a force!. thank you so much
Well explained! I think it should be clarified that the Coriolis effect takes place when air or objects move across latitudes because that is when you get the difference in velocities. If any object (or air) were to move perfectly parallel to a line of latitude (equator or otherwise) there shouldn't be any effect because there is no change in tangential velocity. I would also say that the effect is greater depending on the angle the object moves in relation to the lines of latitude. The closer it is to 90 deg. (or perpendicular to a line of latitude) the more effect will be observed as the tangential velocity reduces more rapidly. As you move toward 0 deg. or closer to being parallel to a line of latitude, the less effect will be observed. To say there is no effect at the equator would only be applicable to the object moving parallel to the equator. If the object is launched perpendicular from the equator (due north for example) then Coriolis effect will occur.
You cannot move parallel to any line of latitude other than the equator without constantly changing direction. So, something launched directly east or west at 40 degrees N latitude will immediately veer off to the south and will remain on a great circle trajectory.
Great video, Sir. Loved the way you confidently spoke. Sir, the video got ended right when you were about to demonstrate how the Coriolis is zero at the equator and maximum at the poles, can you please share that demo video also?
Hello C Hammond. The cannon ball will rotate to the east with the rotating earth (and possibly land on your head).. Keep in mind, the earth does not simply rotate beneath the atmosphere. The atmosphere moves with the rotating earth.
I am confused? I thought to experience the Coriolis effect, two reference frames were needed? For example, throwing a ball in a straight line on a spinning roundabout and as a observer on the roundabout, observe an apparent curve of the ball that when viewed from the non rotating reference frame shows the ball actually traveling in a straight line. If the plane carrying the package is moving in an atmosphere that is in lock step with the earth's rotation, then there would be only reference frame and no Coriolis?
Spin at the equator is equal to spin at the north or south pole. If you lay down on the equator with your head north and your feet to the south, your body is oriented the sane as someone standing on the north pole. In 24 hours, you will spin around once just as someone at the pole spins around.
Great illustration of horizontal coriolis point of impact shift for projectiles launched in north or south direction. What is much harder to illustrate or visualize is the horizontal coriolis point of impact shift that occurs when the same projectile is launched due east or due west. The result is the same regardless of which azimuth direction the projectile is launched. It misses the target to the right in northern hemisphere. misses target to the left in the southern hemisphere even when shooting due east or due west or notheast or northwest.
I pitch Horseshoes. With a distance of 40 feet and a flight time of 2.2 seconds playing in Northern Illinois, how far to the left will my shoes land while facing North?
Just short of 3” , therefore, you aim to the right by 6” and land those suckers spiraling dead center. You sir, will be MVP of the year with the new data.
Airplane pilots don't specifically dial in a course correction for coriolis effect. They're flying thru the body of air which is, more or less, rotating along with the earth.
It's wrong to say that it's not a force. I did a lot of planet trips in a simulator (I know that Coriolis is built in) and apply force to compensate for what forces exist in the beginning and arrive in the planned spot. It's not much compensation needed but it must be considered as a vector force where you turn a little the other way and bleed off the wrong vector you have already when starting out. Logically you will aim at the arrival point and that's the illusion or navigational mistake to compensate for. As you fly the reality of the path is creating a curve (on the ground) which is longer than you think it is from your perspective (as a pilot in this example.) To fly longer or a bit sideways is then a proof of force.
Yes, if the ball was thrown slower to the north, then the amount of Coriolis would be less than a ball thrown faster to the north. A better way to think of this is by examining air in the upper atmosphere. The air is moving quickly, in part due to lack of surface friction (it's up high and not near the ground). Now, compare it to air flowing near the surface. It is going slower, in part due to surface friction. The slower air has less Coriolis and will therefore rotate more with the rotating Earth as opposed to the fast flowing air in the upper atmosphere.
Wrong, slower throw equals more effect. Common sense. If a ball was thrown from Texas to North Dakota(say directly north) slowly it would end up on the west coast. If it is thrown fast it would end up in west Nevada...so it did not go as far west.
If a ball was thrown from Texas (and was able to maintain flight for an extended period of time), it would end up at a longitude EAST of where it started, not west. Re-watch the video (and possibly watch some others), it seems you're missing the fact of the continued momentum in the direction it was going before launch (east) and the difference of that momentum (1,000mph) vs the momentum of its destination latitude (500mph), resulting in a net easterly velocity of 500mph in addition to the northerly velocity.
Good question, Jamie. Keep in mind, that we are a part of the 1000 mph rotation, not exclusive from it. So, if you were traveling in an airplane going to the east at 500 mph, that doesn't mean that the Earth is rotating by you by the difference of 500 mph. Rather, your airplane is moving 500 mph to the east relative to the Earth. Basically, your airplane maintains the 1000 mph Earth rotation plus the 500 mph airplane speed.
using the same concept, it seems that when we fling a rock from the equator southwards, the rock will go right. equator speed=1000mph east, south speed = 500 mph east. so ultimately seems like rock covers 500 miles east in 1 hour. Please clarify the doubt
Think of it this way: If the object is moving toward the north, and you are facing north, then the object is moving to YOUR right (or, to the right of initial motion). This was the example I used in the lecture. Now, if the object is moving south (in the southern hemisphere), and you are facing south, then the object will move to YOUR left (or, to the left of initial motion). Keep in mind, the object will have the initial velocity of 1000 mph to the east. BUT, if you are facing south, then "to the east" will be toward the left of you (whereas if you are facing north, then "to the east" will be toward the right of you).
Ken Yanow You are mistaken in asserting that there is no Coriolis at the equator. While it does not exist for north/south motion, it DOES exist for east/west motion. But it manifests itself in up/down instead of left/right drift. The Coriolis effect is defined as the cross product of two vectors - the earth's angular velocity about its axis and the relative motion of the object (e.g. plane). The cross product's results increase in proportion to the degree that the motion is perpendicular to the axis of rotation and there are TWO perpendicular directions. The mutually orthogonal vector triad is comprised of the axial (earth rotation axis), radial (a vertical axis at the equator), and tangential (east/west axis at the equator). The Coriolis effect exists for both radial and tangential motion.
Ken Yanow The Coriolis force is defined as 2 * Omega cross Vrel. What you're saying then is that motion in the east/west direction at the equator is not perpendicular to the Earth's axis because for the Coriolis effect not to exist, the vector cross product of that motion vector and the Earth's axis of rotation must be zero. I;m sorry to differ with you, but my math skills are good enough that I trust them more than your assertion.
Greg Parrott Hi Greg, I'm not questioning your math. It looks pretty good to me. What I am saying, however, is that the phenomenon you are calculating out is called the Eötvös effect. See en.wikipedia.org/wiki/Eötvös_effect. Here is another link that I suspect you would really enjoy reading: www.lextalus.com/pdf/The%20Coriolis%20Effect.pdf The article discusses exactly what you are describing (mathematical conflicts regarding the Coriolis Effect).
+swayambhu tripathi This is more of a visualisation effect and the explanation attempted is more physical/practical than mathematical. Coriolis is a pseudo force which is applied to account for the Earth's rotating frame of reference. Now to understand this we should first know that 1. When Earth rotates, a certain point or location on earth moves along the latitude of the place or location. 2. If you try or attempt to fly straight from that point or location (independent of earths motion), you'll be following a great circle (that is the circumference of the earth passing through that point, depending on the direction you choose ) and not the latitude of that place. Hence the apparent deflection is the difference of these paths (i.e. the one through which the location is moving that is the latitudinal path and the one which you r following that is the circumference or the great circle) This will now help you understand that as you go towards the poles the angles between the great circle and the latitudes keep on increasing (even if you select a great circle tangential to the latitudes) and therefore we say that Coriolis effect is increasing. Whereas at the equator, the latitude itself carves out a great circle (coriolis effect is zero). We move along the same path which the point or location is following along with the rotating earth.
Help, I am a middle school high school teacher and am struggling with this question. If I am standing on top of the Earth and Ia torpedo toward the equator will the torpedo curve to the east or west from the vantage point of the person on the north pole?
Hi Ken, My daughter asked me this question which i could not answer..... if a Aeroplane is traveling from west to east directly along the equator at 500mph and the spin of the earth is at 1000mph are we actually going to get anywhere?
If an airplane is travelling at 500 mph west to east along the equator, which itself is travelling 1000 mph east, then its net speed relative to the axis of the Earth is 1500 mph to the east. In the reverse direction, its net speed relative to the axis of Earth, is 500 mph to the east. What ultimately matters to us, when defining "get anywhere", is that we change positions relative to the surface of the Earth. There is no such thing as absolute velocity, so it is just as valid to measure speeds relative to the ground that is also moving. You can put this to the test, with what is called the Eotvos effect. Take a sensitive scale (accurate to the gram on for a 500 gram sample), and a reference mass with you, when you fly on a plane. Measure its weight on the ground as control trial, and measure its weight in the plane flying both directions. You should see an apparent increase in its weight when travelling westward, and an apparent decrease in its weight. There is no such thing as absolute velocity, but there is such a thing as absolute rotation. The change in apparent weight of your reference mass, is due to the fact that the plane has to accelerate downward more when travelling east, than it does when parked at the airport. And it accelerates downward less, when travelling west, and your sample will weigh closer to its true weight.
I’m confused as to how 60 degrees is the halfway point of latitude instead of 45 degrees. If the North Pole is 90 degrees away from the equator then wouldn’t the halfway point be 45 degrees? So therefore, it would be 12,000 miles in circumference @ 45 degrees instead of @ 60 degrees.. someone could explain that more because I just don’t see it.. thanks
The formula to calculate radius at different latitudes involves a fair amount of trigonometry. Without going into a bunch of detail, the simple formula (after a lot of derivation) for the circumference at any given latitude is CE x cos (Lat). Here, CE is the circumference of the Earth at the equator, and Lat is the latitude you are seeking to calculate. If you are at a latitude of 60 degrees, then cos60 = 0.5. Therefore, at 60 degrees latitude, the circumference is half the circumference at the equator.
Ken Yanow Ah. I thought it was as simple as just cutting the degree of the variables in half.. the formula you presented helps. I guess it was hard to comprehend because the earth is a somewhat perfect sphere, so I thought it would be directly half. My interest in all this is growing though thanks to your detailed videos!
Ken Yanow okay okay it just hit me. I was totally overlooking the fact that the meridians (in distance) are closing increasingly quick the closer you get to 90 degrees, so the middle would actually be closer to that end, rather than the equator. I had to look at a triangle and cut it in half, horizontally, and then see that the halfway point along that cut would actually have to be closer to the point, to equal out exactly half of the entire length of the base. Visuals help me understand lol. I got it though
Keep in mind, when the airplane takes off, it maintains its momentum it had right before takeoff (i.e., 1000 mph). Once in the air, the forces on the plane are gravity (downward), lift (upward), drag (against the wind), and thrust (forward). After it's initial climb, a plane moves at constant speed with respect to the air. Gravity and lift are in balance, keeping the plane at a constant altitude, and drag and thrust are also in balance, keeping the plane at a constant speed with respect to the air. Our atmosphere moves more or less with the rotating Earth. The plane does, too, plus or minus it's speed with respect to the air. In fact, a plane oftentimes gets better performance flying west to east, in the direction of the planet's rotation. The reason is that the airplane moves with the air. For the most part, the atmosphere does move with the planet. There are some exceptions to this rule. One of the most notable exceptions are the jet streams that move at fairly high speeds from west to east. Airplanes flying to the east will take advantage of jet streams if they can. Airplanes flying to the west, against the Earth's rotation, try to avoid those jet streams if at all possible.
According to you coriolis force is directly proportional to the speed of wind ..Max speed of wind will be near equator but at equator the coriolis force in negligible..... how can this be true ??
Well, it's not according to me! :). Max speed of wind is not necessarily at the Equator. While it's true that the trade winds converge at the Equator, and the trade winds can be fast, that doesn't mean winds at the Equator are max (in fact, due to the convergence of air flow at/near the Equator, i.e., the Intertropical Convergence Zone, winds can be pretty dull. With regards to the speed of wind and the Coriolis, as one goes to altitude, less surface friction means that the air can flow faster (i.e., jetstreams). So, as one goes to altitude, there is a strengthening of the Coriolis effect, helping to create geostrophic flow.
@@jamesaldridge1459 Lol, calling out people's ignorance is a daily task in my line of work. Grow up, admit you're mistaken, and learn what "disrespectful" means instead of crying immediately. All I said is that you can't interchange 2 separate things and then made a quick quip referencing shooting experts that make that same mistake
Hello Frank. Gravity does not change from place-to-place on Earth (any differences are minimal, which will be explained below). The force of gravity equation between two objects is F(g) = Gm(1)m(2)/R^2. In this equation, "G" is the gravitational constant, m(1) is the mass of one object, m(2) is the mass of the second object, and R is the distance between the two objects. So, for any object on Earth, the gravitational pull it feels toward the Earth is dependent upon the mass of the Earth and the distance the object is from the center of the Earth. The mass of the object itself is rather negligible compared to the mass of the planet, so I'm not even considering it in the equation. Because Earth is a sphere, R is the same whether you are at the equator or at the poles. Therefore, gravity is the same everywhere. I alluded to the fact that there are minimal differences to gravity at different places. Those minimal differences could be due to the density of the Earth that is below your feet being more or less than somewhere else (therefore, altering the relative mass of the Earth below you and subsequently changing the gravity relative to somewhere else) or your distance to the center of the planet is slightly more or slightly less than somewhere else (for example, the Earth is almost a perfect sphere, but not quite. The radius of the Earth from the equator to the center of the planet is slightly longer than the radius of the Earth from the pole to the center of the planet. Because of this, the gravity at the equator would be slightly less than that at the poles -- remember, force of gravity is inversely proportional to R. If R is longer, then F(g) is less.) But, really, whether we are talking about slight changes in m(1) or slight changes in R, we are talking about a fraction of a percentage. The difference is really, really small. The Coriolis is not a gravity phenomenon. Rather, it is simply the result of a rotating planet. Thanks for the question.
you forgot gravity? and the atmosphere? the clouds don't leave us behind and end up in india, considering that the earth moves at 444m/s at the equator
why are you changing the relative positions of two points, which in your case are Ecuador and Nova Scotia. You are showing that after 1 hour of rotation Ecuador is 500 miles ahead of Nova Scotia. And as the package is going more and more towards the north its drifting towards the east because of the linear velocity of earth but the package is in the earth's atmosphere and the speed it gained is because of earth's velocity therefore its speed should come down as the velocity of earth near your second point is also less. So confusing.
Try not to think of it as Ecuador (EC) moving ahead of Nova Scotia (NS). If the world were flat, and EC were to move east at a rate twice that of NS, then yes, it would move ahead of it; but then, if the world were flat, this would all be moot. The fact is, the world is round, hence the rotation which creates this effect. So EC has to travel twice the distance (Earth's circumference) in the same amount of time as NS, resulting in traveling twice the speed. They're always going to stay at the same longitude, because they're locked in place, but this means that EC, having a longer distance to cover in the same time, has to move faster. Hence the 1,000 mph easterly velocity there vs the 500 mph easterly velocity at NS. Next, remember than any object launched at EC starts out, before launch, with a velocity of 1,000 mph east. In other words, it's equal to that of EC, which is why it stays in place. If you throw a ball straight up into the air, it comes straight back down; it doesn't fly to the west because you're on the ground moving at 1,000 mph to the east. The ball has the same easterly velocity as EC and you, and you're only adding a vertical momentum to it, which gravity then works against and brings it back to you. Now, if you take that same ball and throw it north fast enough for it to resist gravity long enough to make it to NS, it's moving 1,000 mph east plus whatever northerly velocity you gave it when you threw it (let's say NS is 1,000 miles north of EC and you throw it at 1,000 mph, so it'll take an hour to travel the distance required). Meanwhile, NS is moving east at a mere 500 mph (remember, it has less distance to travel in the same 24 hours to end up back in the same spot, because of the smaller circumference at that point). If, as stipulated, it takes the ball an hour to travel the distance between EC and NS, after that time it will have traveled 1,000 miles north and 1,000 miles east (remember, it had a velocity of 1,000 mph in each direction). Unfortunately, NS has only traveled 500 miles east, so the ball will end up 500 miles east of it. EC and NS are still at the same longitude of each other, though. The confusion you're having is related to his representation of this 3D, spherical movement on a 2D, flat drawing. What you have to keep in mind is that while NS is moving half the distance, because it's at a higher latitude and so doesn't have to move as far to complete a full rotation, it still keeps up with EC despite EC's faster movement. Consider a tire and wheel: the outer edge (where the tread is) has to move at a faster rate to complete a rotation compared to the lugnuts. In a given time, a given point on the outer part of the tire will travel a farther linear distance than the lugnut between it and the axle, but they will always remain in line. If the Earth were a rotating cylinder, this effect wouldn't happen, because all points would rotate at the same speed. Being a sphere, though, means different locations move at different speeds. If you could take his drawing and wrap it around a ball (I say 'if you could' because you can't wrap a flat drawing around a sphere or, inversely, unwrap a sphere into a flat drawing, hence adjustments that are made to maps to compensate for this), you would find that the two would no longer be out of alignment. It might help to look at it from a different angle, looking down on the Earth from above the North pole (you could visualize this with a globe or even a ball). EC: NS: | |_ |___ |_ | |_ |___ |_ | |___ You can see that, looking at it from this angle, EC has to travel much further to end up at the same longitude as NS after this period of time, hence its higher speed. As counter-intuitive as all of this seems, the basic concept you have to remember for it to make sense is that EC is moving faster and, because of this, anything that is "launched" from there (ball, cannonball, plane, wind) starts out with that higher easterly velocity and so will move east faster than things (land, people, air/wind) at a more northerly latitude. While his explanation does well to describe the reason for the effect, I can certainly see how it would cause confusion, because it's not really that the slower movement of NS is causing it to not be at the same longitude of EC, but rather that the faster speed of the object imparted on it by the faster speed of EC causes it to circle the globe faster than NS and therefore get ahead of it. One final way of looking at it (I find that looking at things from multiple angles helps in understanding them, and some people may not get it from some explanations but things just click from another) is to draw it out from the north pole viewpoint. Draw two concentric circles, one much smaller than the other, like a donut. Now put an x on each one at the 5 & 7 o'clock positions. You'll see that the two 5 o'clock (right) x's are lined up with each other and the two 7 o'clock (left) x's are also lined up with each other, however, the two on the inner circle (NS) are very close together compared to the two on the outer circle (EC). Now imagine launching an object from the left EC x toward the left NS x. Again, assuming it will take an hour to get from the outer circle (the southern latitude where EC is located) to the inner circle (the northern latitude where NS is located), and assuming it takes an hour for the speed of the Earth's rotation to cause EC and NS to get from each left x (their starting longitude) to the right x's, you can see the easterly velocity imparted on the object by the velocity of EC will cause it to move east, in that hour, by a distance significantly greater than that traversed by NS in that same time. This means that as it moves north toward NS, it's higher easterly velocity will carry it much farther than the distance necessary to end up in NS. It might help to draw an arrow from the left x to the right x on each latitude line (each circle), and then draw an arrow the length of the one on the outer circle on the inner circle starting at the left x, and you'll see that distance carries it well past the location of NS at the end of that hour.
If you throw the package from the northern hemisphere, the package will veer to the right of initial motin. If you throw the package in the Southern Hemisphere, the package will veer to the left of initial motion.
***** Hi Chris, this is really a visualization game. Imagine that you are on some latitude (30 degrees north, for example), and you shoot a rocket due east. That rocket is going in a straight line relative to a rotating Earth. Now imagine that there is a stationary tree at 30 degrees north. The tree that is moving east with the Earth is not moving in a straight line. Rather, because it is moving with the Earth on a rotating sphere, it is making an arc motion. So, let's say that this rocket you shot is approaching the tree. The rocket is moving straight, but the tree is moving in an arc. By the time the rocket reaches the point of the tree, the tree has moved, essentially to the left of the moving rocket (that is, the rocket has appeared to veer to the right of the tree).
If my cannon was sitting on the equator and I then shot a cannon ball straight up into the air, would the cannon ball land on the west side of the cannon? Since the Earth is rotating to the east at 1000 mph, this is what I would expect to happen. If the ball doesn't land on the west side of the cannon, what would that mean?
Alastair Douglas Yes,if ya shot one hundred cannon balls I'm sure at least a couple would come back down to dead center. But that would disprove the Coriolis Effect. Would it not?
+C Hammond I found this interesting and decided to do the calculation. For a simplified example (neglecting air resistance and assuming constant gravitational acceleration) it turns out that a cannon ball fired straight up with a velocity of 1000 feet per second would land about 94 feet to the west of where it was fired.
In the Northern Hemisphere, drift will be to the right of initial motion. In the Southern Hemisphere, drift will be to the left of initial motion. Using "east" and "west" is probably not the best way to express this. For example, in the northern hemisphere, if an object is traveling south, it will get veered to the right (with would be east). But, if it is traveling north, it is getting veered to the right, but in this case, that would be west.
@@kenyanow8274 Perhaps I'm confused but, does not the given argument show that if you stand on the equator and pitch an object towards either pole in the direction of a longitude line then it will land to the East of that longitude line.
Peter Higgins, yes, but what if you're in the southern hemisphere at around 60 degrees south latitude and heave the object toward the equator? It will get veered to the left of initial motion (in this case, West).
So if guns and cannons have to account for the "coriolis effect" then why does the earth traveling through space at 67,000 mph not effect the bullet and only the 1000 mph rotation of the earth does?
Because the entire Earth is moving at the same 67,000 mph, so both the gun/cannon and its target are moving at that same speed, and therefore the bullet/cannonball has that velocity plus the velocity from being launched from the gun/cannon. Conversely, different points of the Earth are moving at different speeds due to rotation, so the gun/cannon and its target are moving at slightly different speeds from this, thereby requiring adjustment.
Because all of Earth's orbital motion is caused by gravity, and the Earth is in free fall around the sun. Ignoring the sun's tidal forces, there is no net constraint force necessary to keep you in the same position on Earth, against the sun's gravity, to keep you orbiting with the Earth. The sun's gravity is entirely "used up" in causing this motion, and thus both the orbital acceleration of Earth and the sun's gravity of Earth, will nullify each other in the reference frame of the Earth. Also, that 67,000 mph orbital speed may seem high, but relative to the radius of curvature of the Earth's orbit, it is a hell of a lot less than the angular speed of the Earth's rotation. Angular speed matters more for the Coriolis effect than tangential speed.
So if a plane travels from West to East it travels in the same direction of the rotation of the globe whereas if the plane travels from East to West it travels against the direction of the rotation of the globe. 🤔 Let me enlighten you: Earth is not a rotating globe guys!
So, basically, the rate of change of the hypothetical object's velocity is part of a function of the cosine of the distance from the equator, as the equator is the origin for the Coriolis effect.. Okay. Nice explanation!
the wording was a bit weird to start. "The circumference of the Eart at [x]," never reaches zero, but a horizontal slice perpendicular to the axis of rotation could reach zero.Try not to confuse more people than you need to
2:37-wtf?! Tell me: is there a difference between what a person physically feels closer to the axis on a merry-go-round than a person farther out? The answer is yes.
@Flat Earth Data no. Can you demonstrate it for yourself? Go put a pendulum on a lazy susan with the bob anywhere but the middle. Spin it. Why does the bob deflect to the outside? Answer: because it's accelerating INWARDS towards the centre.
@Flat Earth Data no...a lazy susan isn't the Earth, but it can be used to demonstrate centrpetal acceleration. Didn't you do this in grade 11 or 12 in physics like the rest of Canada?
@Flat Earth Data Earth doesn't need to be a perfect circle. Neither does the lazy susan. The path the pendulum will make is a perfect circle, assuming proper construction of the lazy susan. I don't need to find the COM in this example, but if the massive bob is a sphere of uniform material, it's quite easy. Do you need me to explain that?
I was looking for a mathematics explanation of the Coriolis Effect, like solving Newton’s 2nd law of motion in a rotating frame of reference, possible using Lagrange equation to make it easier, and I found this crap....this video is just about relative motion and even so: Totally wrong! 😢
I’m curious, Erik. What do you suppose is incorrect? I’ll also refer you to the following site for a similar explanation: www.nationalgeographic.org/encyclopedia/coriolis-effect/ If you’re looking for a detail derivation of the physics, this video is not for that (and clearly not for you). It is a concept based introduction to the topic.
This is awful. I'm a musician but am aware of how wrong this is. If you put two pins on a globe in the two locations he mentions, perfectly in line north to south, and tie a string to them, and then spin the globe the string always stays straight. Yes, earth spins faster at equator but it's traveling a greater distance! So the two points remain in line regardless. This is silly. According to his logic a helicopter could fly perfectly north from the equator and end up hundreds of miles east and the passengers could literally watch the world spinning under them. Nonsense.
This has already been explained multiple times here as well as elsewhere, which makes me doubt you actually want an answer and suspect that, more likely, you just want to insist that the Earth can't be round and spinning. It's not a case of a plane going ~150mph (approximate landing speed) landing on a surface spinning at 1,000mph, it's a case of a plane moving at ~1,150mph landing on a surface spinning at 1,000mph. This is because the plane always has the initial velocity of the spinning Earth *in addition to* its own velocity.
the pane & air is spinning along with the ball... just like how you would be able to maneuver and hover a drone inside a plane that you know is traveling 600 m/hr (traveling at a consistent speed, like the earth is rotating AND ALSO flying through space making a big circle around the sun) speeding up or slowing down makes g force
I’m not sure what you’re talking about. Inside the Earth’s environment? You mean anywhere from the top of the atmosphere to sea surface? Here is another resource that discusses how Coriolis changes with elevation. geography.name/how-does-the-coriolis-effect-influence-wind-direction-at-different-heights/
@@kenyanow8274 I am glad you replied. Your explanation is comparing the Earth to a Merry-go-round, isn't it? But unlike the Merry-go-round if we make a jump from the Earth's surface we will not land at a different place viz. the Earth doesn't pass under our feet. Because the Earth is a stationary for everything that is inside its atmosphere. Plz clarify...
@@binod3386 That might be a different video :). No, I did not refer to a merry-go-round. But, a merry-go-round analogy is not a bad one. In the analogy, no one is jumping off the merry-go-round. The Coriolis is the result of objects moving within a rotating reference frame.
Your explanation makes no sense! The rate of rotation around the axis of the earth is constant no matter where on the surface u are standing. If Ecuador is due South of Nova Scotia, and u have a zip line that follows the trajectory of ur sling shot package, and the package is attached to the zip line, the package would end up in Nova Scotia after an hour. Your suggesting that if I jump at the equator, I will be slammed into the wall that is opposite the direction of the earth’s rotation, and if i do so in Nova Scotia then the distance to the wall that i would slam into could be half as far from the point i jump from, cause i’m moving slower? No wonder american kids suck at science and math! If their teachers can’t figure it out! Gravity is pulling everything down at a constant, that doesn’t change just because an item gets launched into the air... and the speed someones traveling at the equator would appear the exact same as someones speed standing at 60 degrees North Latitude relative to an observer in space because the person at 60 degrees latitude would appear on one horizon and disappear on the other horizon at the same times as the person standing at the equator, as a 2 dimensional view, they are moving at the same speed, but your suggesting that somehow the space observer can comprehend the depth distinction due to curvature, and i don’t think any one can easily distinguish the difference 2 dimensionally from observing in 3 dimensions of a sphere.
5:32-wtf?! “Apparently of initial motion????” Is the classroom empty? Is no one actually listening to this mong??? 6:05-“If everything is left undone???” Who taught you syntax and grammar?? No one accounts for the rotation of the earth when flying. If they did, then that would mean the earth is MOVING under the package. How does this numpty change the direct line from Quito to NS to an offset of 500 miles? They were in line before and now they’re offset? FFS. HEAD toward NS and you’ll get there. FFS.
What a bunch of garbage. The spin of the earth changes the trajectory of an object moving directly north apparently 45°? How then do we explain an airplane flight from LA to New York taking more time than from New York to LA nonstop? If Coriolis is real then we should be able to go straight up in a helicopter, hover for a few hours and come straight back down and land in a completely different area but that does not happen.
![เปิด อาณาจักร หมื่นล้าน "สาระตั้ม" l [Nickynachat]](http://i.ytimg.com/vi/3XE0nPF4YkQ/mqdefault.jpg)
This is the best explanation on the internet. he explains why we don;t have coriolis force at equator, why the coriolis force is to your right in NH, and left in SH. also explain clearly that the coriolis force is not a force!. thank you so much
Well explained! I think it should be clarified that the Coriolis effect takes place when air or objects move across latitudes because that is when you get the difference in velocities. If any object (or air) were to move perfectly parallel to a line of latitude (equator or otherwise) there shouldn't be any effect because there is no change in tangential velocity. I would also say that the effect is greater depending on the angle the object moves in relation to the lines of latitude. The closer it is to 90 deg. (or perpendicular to a line of latitude) the more effect will be observed as the tangential velocity reduces more rapidly. As you move toward 0 deg. or closer to being parallel to a line of latitude, the less effect will be observed. To say there is no effect at the equator would only be applicable to the object moving parallel to the equator. If the object is launched perpendicular from the equator (due north for example) then Coriolis effect will occur.
You cannot move parallel to any line of latitude other than the equator without constantly changing direction. So, something launched directly east or west at 40 degrees N latitude will immediately veer off to the south and will remain on a great circle trajectory.
Great video, Sir. Loved the way you confidently spoke. Sir, the video got ended right when you were about to demonstrate how the Coriolis is zero at the equator and maximum at the poles, can you please share that demo video also?
Yes, please share that, too!
This is the best explanation I have seen that I can understand. Better than fancy color animations which do me nothing.
Very cool lecture. Well explained, even to a 50 yr. old. I know this effect but explained it better than any written text.
I love this, Ken!
Thanks, easy to understand for science guys like me who grow up with vectors
an explanation that makes everything makes sense
Hello C Hammond.
The cannon ball will rotate to the east with the rotating earth (and possibly land on your head).. Keep in mind, the earth does not simply rotate beneath the atmosphere. The atmosphere moves with the rotating earth.
Show that in an experiment. Show atmosphere sticking to a spinning ball
I am confused? I thought to experience the Coriolis effect, two reference frames were needed? For example, throwing a ball in a straight line on a spinning roundabout and as a observer on the roundabout, observe an apparent curve of the ball that when viewed from the non rotating reference frame shows the ball actually traveling in a straight line.
If the plane carrying the package is moving in an atmosphere that is in lock step with the earth's rotation, then there would be only reference frame and no Coriolis?
Thank you very Much from france, you are realy the best, more than everybody! Bravo!
Spin at the equator is equal to spin at the north or south pole. If you lay down on the equator with your head north and your feet to the south, your body is oriented the sane as someone standing on the north pole. In 24 hours, you will spin around once just as someone at the pole spins around.
Great illustration of horizontal coriolis point of impact shift for projectiles launched in north or south direction. What is much harder to illustrate or visualize is the horizontal coriolis point of impact shift that occurs when the same projectile is launched due east or due west. The result is the same regardless of which azimuth direction the projectile is launched. It misses the target to the right in northern hemisphere. misses target to the left in the southern hemisphere even when shooting due east or due west or notheast or northwest.
It doesn't. We checked.
I pitch Horseshoes.
With a distance of 40 feet and a flight time of 2.2 seconds playing in Northern Illinois, how far to the left will my shoes land while facing North?
Just short of 3” , therefore, you aim to the right by 6” and land those suckers spiraling dead center. You sir, will be MVP of the year with the new data.
where is the reamining lecture?
Airplane pilots don't specifically dial in a course correction for coriolis effect. They're flying thru the body of air which is, more or less, rotating along with the earth.
It's wrong to say that it's not a force. I did a lot of planet trips in a simulator (I know that Coriolis is built in) and apply force to compensate for what forces exist in the beginning and arrive in the planned spot. It's not much compensation needed but it must be considered as a vector force where you turn a little the other way and bleed off the wrong vector you have already when starting out. Logically you will aim at the arrival point and that's the illusion or navigational mistake to compensate for. As you fly the reality of the path is creating a curve (on the ground) which is longer than you think it is from your perspective (as a pilot in this example.) To fly longer or a bit sideways is then a proof of force.
can you explain once more why Coriolis force is max at poles and zero at equator. I can't understand your twist and turn reason sir
See www.nationalgeographic.org/encyclopedia/coriolis-effect/
Yes, if the ball was thrown slower to the north, then the amount of Coriolis would be less than a ball thrown faster to the north. A better way to think of this is by examining air in the upper atmosphere. The air is moving quickly, in part due to lack of surface friction (it's up high and not near the ground). Now, compare it to air flowing near the surface. It is going slower, in part due to surface friction. The slower air has less Coriolis and will therefore rotate more with the rotating Earth as opposed to the fast flowing air in the upper atmosphere.
Wrong, slower throw equals more effect. Common sense. If a ball was thrown from Texas to North Dakota(say directly north) slowly it would end up on the west coast. If it is thrown fast it would end up in west Nevada...so it did not go as far west.
If a ball was thrown from Texas (and was able to maintain flight for an extended period of time), it would end up at a longitude EAST of where it started, not west. Re-watch the video (and possibly watch some others), it seems you're missing the fact of the continued momentum in the direction it was going before launch (east) and the difference of that momentum (1,000mph) vs the momentum of its destination latitude (500mph), resulting in a net easterly velocity of 500mph in addition to the northerly velocity.
Awesome explanation
Good question, Jamie. Keep in mind, that we are a part of the 1000 mph rotation, not exclusive from it. So, if you were traveling in an airplane going to the east at 500 mph, that doesn't mean that the Earth is rotating by you by the difference of 500 mph. Rather, your airplane is moving 500 mph to the east relative to the Earth. Basically, your airplane maintains the 1000 mph Earth rotation plus the 500 mph airplane speed.
Very well explained
using the same concept, it seems that when we fling a rock from the equator southwards, the rock will go right. equator speed=1000mph east, south speed = 500 mph east. so ultimately seems like rock covers 500 miles east in 1 hour. Please clarify the doubt
Think of it this way: If the object is moving toward the north, and you are facing north, then the object is moving to YOUR right (or, to the right of initial motion). This was the example I used in the lecture. Now, if the object is moving south (in the southern hemisphere), and you are facing south, then the object will move to YOUR left (or, to the left of initial motion). Keep in mind, the object will have the initial velocity of 1000 mph to the east. BUT, if you are facing south, then "to the east" will be toward the left of you (whereas if you are facing north, then "to the east" will be toward the right of you).
Ken Yanow
You are mistaken in asserting that there is no Coriolis at the equator. While it does not exist for north/south motion, it DOES exist for east/west motion. But it manifests itself in up/down instead of left/right drift. The Coriolis effect is defined as the cross product of two vectors - the earth's angular velocity about its axis and the relative motion of the object (e.g. plane). The cross product's results increase in proportion to the degree that the motion is perpendicular to the axis of rotation and there are TWO perpendicular directions. The mutually orthogonal vector triad is comprised of the axial (earth rotation axis), radial (a vertical axis at the equator), and tangential (east/west axis at the equator). The Coriolis effect exists for both radial and tangential motion.
Greg Parrott
What you are referring to is the Eötvös effect. At the Equator, the Coriolis Force is 0.
Ken Yanow
The Coriolis force is defined as 2 * Omega cross Vrel. What you're saying then is that motion in the east/west direction at the equator is not perpendicular to the Earth's axis because for the Coriolis effect not to exist, the vector cross product of that motion vector and the Earth's axis of rotation must be zero. I;m sorry to differ with you, but my math skills are good enough that I trust them more than your assertion.
Greg Parrott
Hi Greg,
I'm not questioning your math. It looks pretty good to me. What I am saying, however, is that the phenomenon you are calculating out is called the Eötvös effect.
See en.wikipedia.org/wiki/Eötvös_effect.
Here is another link that I suspect you would really enjoy reading:
www.lextalus.com/pdf/The%20Coriolis%20Effect.pdf
The article discusses exactly what you are describing (mathematical conflicts regarding the Coriolis Effect).
Can you please explain how this effect applies to aeroplanes.
If an object was to move parallel (or perpendicular to the axe) to the rotation of the earth ... would it still be deflected?
+swayambhu tripathi
This is more of a visualisation effect and the explanation attempted is more physical/practical than mathematical.
Coriolis is a pseudo force which is applied to account for the Earth's rotating frame of reference. Now to understand this we should first know that
1. When Earth rotates, a certain point or location on earth moves along the latitude of the place or location.
2. If you try or attempt to fly straight from that point or location (independent of earths motion), you'll be following a great circle (that is the circumference of the earth passing through that point, depending on the direction you choose ) and not the latitude of that place.
Hence the apparent deflection is the difference of these paths (i.e. the one through which the location is moving that is the latitudinal path and the one which you r following that is the circumference or the great circle)
This will now help you understand that as you go towards the poles the angles between the great circle and the latitudes keep on increasing (even if you select a great circle tangential to the latitudes) and therefore we say that Coriolis effect is increasing. Whereas at the equator, the latitude itself carves out a great circle (coriolis effect is zero). We move along the same path which the point or location is following along with the rotating earth.
+Prekaxl Rose yes that's why it is zero at equator and maximum at poles...right...thanks man...;)
Help, I am a middle school high school teacher and am struggling with this question. If I am standing on top of the Earth and Ia torpedo toward the equator will the torpedo curve to the east or west from the vantage point of the person on the north pole?
Towards the east.
Hi Ken, My daughter asked me this question which i could not answer..... if a Aeroplane is traveling from west to east directly along the equator at 500mph and the spin of the earth is at 1000mph are we actually going to get anywhere?
If an airplane is travelling at 500 mph west to east along the equator, which itself is travelling 1000 mph east, then its net speed relative to the axis of the Earth is 1500 mph to the east. In the reverse direction, its net speed relative to the axis of Earth, is 500 mph to the east. What ultimately matters to us, when defining "get anywhere", is that we change positions relative to the surface of the Earth. There is no such thing as absolute velocity, so it is just as valid to measure speeds relative to the ground that is also moving.
You can put this to the test, with what is called the Eotvos effect. Take a sensitive scale (accurate to the gram on for a 500 gram sample), and a reference mass with you, when you fly on a plane. Measure its weight on the ground as control trial, and measure its weight in the plane flying both directions. You should see an apparent increase in its weight when travelling westward, and an apparent decrease in its weight. There is no such thing as absolute velocity, but there is such a thing as absolute rotation. The change in apparent weight of your reference mass, is due to the fact that the plane has to accelerate downward more when travelling east, than it does when parked at the airport. And it accelerates downward less, when travelling west, and your sample will weigh closer to its true weight.
I’m confused as to how 60 degrees is the halfway point of latitude instead of 45 degrees. If the North Pole is 90 degrees away from the equator then wouldn’t the halfway point be 45 degrees? So therefore, it would be 12,000 miles in circumference @ 45 degrees instead of @ 60 degrees.. someone could explain that more because I just don’t see it.. thanks
The formula to calculate radius at different latitudes involves a fair amount of trigonometry. Without going into a bunch of detail, the simple formula (after a lot of derivation) for the circumference at any given latitude is CE x cos (Lat). Here, CE is the circumference of the Earth at the equator, and Lat is the latitude you are seeking to calculate. If you are at a latitude of 60 degrees, then cos60 = 0.5. Therefore, at 60 degrees latitude, the circumference is half the circumference at the equator.
Ken Yanow Ah. I thought it was as simple as just cutting the degree of the variables in half.. the formula you presented helps. I guess it was hard to comprehend because the earth is a somewhat perfect sphere, so I thought it would be directly half. My interest in all this is growing though thanks to your detailed videos!
Ken Yanow okay okay it just hit me. I was totally overlooking the fact that the meridians (in distance) are closing increasingly quick the closer you get to 90 degrees, so the middle would actually be closer to that end, rather than the equator. I had to look at a triangle and cut it in half, horizontally, and then see that the halfway point along that cut would actually have to be closer to the point, to equal out exactly half of the entire length of the base. Visuals help me understand lol. I got it though
Circumference abt the earth stay same since it's the same sphere no matter your position, changes are the C arc abt the axis.
Excellent 👍
can u demostarate the coroilis force zero at equator and max at poles
Keep in mind, when the airplane takes off, it maintains its momentum it had right before takeoff (i.e., 1000 mph). Once in the air, the forces on the plane are gravity (downward), lift (upward), drag (against the wind), and thrust (forward). After it's initial climb, a plane moves at constant speed with respect to the air. Gravity and lift are in balance, keeping the plane at a constant altitude, and drag and thrust are also in balance, keeping the plane at a constant speed with respect to the air.
Our atmosphere moves more or less with the rotating Earth. The plane does, too, plus or minus it's speed with respect to the air.
In fact, a plane oftentimes gets better performance flying west to east, in the direction of the planet's rotation. The reason is that the airplane moves with the air. For the most part, the atmosphere does move with the planet. There are some exceptions to this rule. One of the most notable exceptions are the jet streams that move at fairly high speeds from west to east. Airplanes flying to the east will take advantage of jet streams if they can. Airplanes flying to the west, against the Earth's rotation, try to avoid those jet streams if at all possible.
This guy’s a fantasist
Very well explained...
super , j'ai enfin compris le principe :)
Et faut biensur que ce soit en anglais qu'on le comprenne... Pourtant j'ai beaucoup cherché dans la langue de nos mères ;)
According to you coriolis force is directly proportional to the speed of wind ..Max speed of wind will be near equator but at equator the coriolis force in negligible..... how can this be true ??
Well, it's not according to me! :). Max speed of wind is not necessarily at the Equator. While it's true that the trade winds converge at the Equator, and the trade winds can be fast, that doesn't mean winds at the Equator are max (in fact, due to the convergence of air flow at/near the Equator, i.e., the Intertropical Convergence Zone, winds can be pretty dull. With regards to the speed of wind and the Coriolis, as one goes to altitude, less surface friction means that the air can flow faster (i.e., jetstreams). So, as one goes to altitude, there is a strengthening of the Coriolis effect, helping to create geostrophic flow.
But there would be coriolis affect on a bullet shooting from west to east and vice versa.
No. That's not Coriolis effect on a bullet. It's eotvos effect. Centrifugal force, not spin.
There absolutely is a vertical coriolis when shooting east/west...
@@jamesaldridge1459 That is not coriolis. It is eotvos. They are not interchangeable no matter what your shooting buddy told you
@@cryangallegos Disrespectful right off the bat, huh? Ok sir. Have a good day. I bet you're more careful with youre wording in person...
@@jamesaldridge1459 Lol, calling out people's ignorance is a daily task in my line of work. Grow up, admit you're mistaken, and learn what "disrespectful" means instead of crying immediately. All I said is that you can't interchange 2 separate things and then made a quick quip referencing shooting experts that make that same mistake
Very good lecture.
So the closer you are to the equator the stronger the gravity of the earth is?
Hello Frank. Gravity does not change from place-to-place on Earth (any differences are minimal, which will be explained below). The force of gravity equation between two objects is F(g) = Gm(1)m(2)/R^2. In this equation, "G" is the gravitational constant, m(1) is the mass of one object, m(2) is the mass of the second object, and R is the distance between the two objects. So, for any object on Earth, the gravitational pull it feels toward the Earth is dependent upon the mass of the Earth and the distance the object is from the center of the Earth. The mass of the object itself is rather negligible compared to the mass of the planet, so I'm not even considering it in the equation. Because Earth is a sphere, R is the same whether you are at the equator or at the poles. Therefore, gravity is the same everywhere. I alluded to the fact that there are minimal differences to gravity at different places. Those minimal differences could be due to the density of the Earth that is below your feet being more or less than somewhere else (therefore, altering the relative mass of the Earth below you and subsequently changing the gravity relative to somewhere else) or your distance to the center of the planet is slightly more or slightly less than somewhere else (for example, the Earth is almost a perfect sphere, but not quite. The radius of the Earth from the equator to the center of the planet is slightly longer than the radius of the Earth from the pole to the center of the planet. Because of this, the gravity at the equator would be slightly less than that at the poles -- remember, force of gravity is inversely proportional to R. If R is longer, then F(g) is less.) But, really, whether we are talking about slight changes in m(1) or slight changes in R, we are talking about a fraction of a percentage. The difference is really, really small. The Coriolis is not a gravity phenomenon. Rather, it is simply the result of a rotating planet. Thanks for the question.
niiiiice, this explanation made sense to me
Please explain what is this 24000 &12000
1 full rotation of Earth is 24 hours (day and night) we travel at 1000 mph Fe, we have travelled 24000 miles at the equator.
you forgot gravity? and the atmosphere? the clouds don't leave us behind and end up in india, considering that the earth moves at 444m/s at the equator
why are you changing the relative positions of two points, which in your case are Ecuador and Nova Scotia. You are showing that after 1 hour of rotation Ecuador is 500 miles ahead of Nova Scotia.
And as the package is going more and more towards the north its drifting towards the east because of the linear velocity of earth but the package is in the earth's atmosphere and the speed it gained is because of earth's velocity therefore its speed should come down as the velocity of earth near your second point is also less.
So confusing.
Try not to think of it as Ecuador (EC) moving ahead of Nova Scotia (NS). If the world were flat, and EC were to move east at a rate twice that of NS, then yes, it would move ahead of it; but then, if the world were flat, this would all be moot. The fact is, the world is round, hence the rotation which creates this effect. So EC has to travel twice the distance (Earth's circumference) in the same amount of time as NS, resulting in traveling twice the speed. They're always going to stay at the same longitude, because they're locked in place, but this means that EC, having a longer distance to cover in the same time, has to move faster. Hence the 1,000 mph easterly velocity there vs the 500 mph easterly velocity at NS.
Next, remember than any object launched at EC starts out, before launch, with a velocity of 1,000 mph east. In other words, it's equal to that of EC, which is why it stays in place. If you throw a ball straight up into the air, it comes straight back down; it doesn't fly to the west because you're on the ground moving at 1,000 mph to the east. The ball has the same easterly velocity as EC and you, and you're only adding a vertical momentum to it, which gravity then works against and brings it back to you. Now, if you take that same ball and throw it north fast enough for it to resist gravity long enough to make it to NS, it's moving 1,000 mph east plus whatever northerly velocity you gave it when you threw it (let's say NS is 1,000 miles north of EC and you throw it at 1,000 mph, so it'll take an hour to travel the distance required). Meanwhile, NS is moving east at a mere 500 mph (remember, it has less distance to travel in the same 24 hours to end up back in the same spot, because of the smaller circumference at that point). If, as stipulated, it takes the ball an hour to travel the distance between EC and NS, after that time it will have traveled 1,000 miles north and 1,000 miles east (remember, it had a velocity of 1,000 mph in each direction). Unfortunately, NS has only traveled 500 miles east, so the ball will end up 500 miles east of it.
EC and NS are still at the same longitude of each other, though. The confusion you're having is related to his representation of this 3D, spherical movement on a 2D, flat drawing. What you have to keep in mind is that while NS is moving half the distance, because it's at a higher latitude and so doesn't have to move as far to complete a full rotation, it still keeps up with EC despite EC's faster movement. Consider a tire and wheel: the outer edge (where the tread is) has to move at a faster rate to complete a rotation compared to the lugnuts. In a given time, a given point on the outer part of the tire will travel a farther linear distance than the lugnut between it and the axle, but they will always remain in line. If the Earth were a rotating cylinder, this effect wouldn't happen, because all points would rotate at the same speed. Being a sphere, though, means different locations move at different speeds. If you could take his drawing and wrap it around a ball (I say 'if you could' because you can't wrap a flat drawing around a sphere or, inversely, unwrap a sphere into a flat drawing, hence adjustments that are made to maps to compensate for this), you would find that the two would no longer be out of alignment. It might help to look at it from a different angle, looking down on the Earth from above the North pole (you could visualize this with a globe or even a ball).
EC: NS:
| |_
|___ |_
| |_
|___ |_
|
|___
You can see that, looking at it from this angle, EC has to travel much further to end up at the same longitude as NS after this period of time, hence its higher speed. As counter-intuitive as all of this seems, the basic concept you have to remember for it to make sense is that EC is moving faster and, because of this, anything that is "launched" from there (ball, cannonball, plane, wind) starts out with that higher easterly velocity and so will move east faster than things (land, people, air/wind) at a more northerly latitude. While his explanation does well to describe the reason for the effect, I can certainly see how it would cause confusion, because it's not really that the slower movement of NS is causing it to not be at the same longitude of EC, but rather that the faster speed of the object imparted on it by the faster speed of EC causes it to circle the globe faster than NS and therefore get ahead of it.
One final way of looking at it (I find that looking at things from multiple angles helps in understanding them, and some people may not get it from some explanations but things just click from another) is to draw it out from the north pole viewpoint. Draw two concentric circles, one much smaller than the other, like a donut. Now put an x on each one at the 5 & 7 o'clock positions. You'll see that the two 5 o'clock (right) x's are lined up with each other and the two 7 o'clock (left) x's are also lined up with each other, however, the two on the inner circle (NS) are very close together compared to the two on the outer circle (EC). Now imagine launching an object from the left EC x toward the left NS x. Again, assuming it will take an hour to get from the outer circle (the southern latitude where EC is located) to the inner circle (the northern latitude where NS is located), and assuming it takes an hour for the speed of the Earth's rotation to cause EC and NS to get from each left x (their starting longitude) to the right x's, you can see the easterly velocity imparted on the object by the velocity of EC will cause it to move east, in that hour, by a distance significantly greater than that traversed by NS in that same time. This means that as it moves north toward NS, it's higher easterly velocity will carry it much farther than the distance necessary to end up in NS. It might help to draw an arrow from the left x to the right x on each latitude line (each circle), and then draw an arrow the length of the one on the outer circle on the inner circle starting at the left x, and you'll see that distance carries it well past the location of NS at the end of that hour.
Thankyou for taking the time to answer this 👍
What if i throw my package eastwards and/or westwards at mid/high latitudes?
If you throw the package from the northern hemisphere, the package will veer to the right of initial motin. If you throw the package in the Southern Hemisphere, the package will veer to the left of initial motion.
***** Hi Chris, this is really a visualization game. Imagine that you are on some latitude (30 degrees north, for example), and you shoot a rocket due east. That rocket is going in a straight line relative to a rotating Earth. Now imagine that there is a stationary tree at 30 degrees north. The tree that is moving east with the Earth is not moving in a straight line. Rather, because it is moving with the Earth on a rotating sphere, it is making an arc motion. So, let's say that this rocket you shot is approaching the tree. The rocket is moving straight, but the tree is moving in an arc. By the time the rocket reaches the point of the tree, the tree has moved, essentially to the left of the moving rocket (that is, the rocket has appeared to veer to the right of the tree).
But air will be faster at equated not on poles so coriolis should be greater on equator
The Coriolis effect has nothing to do with air.
If my cannon was sitting on the equator and I then shot a cannon ball straight up into the air, would the cannon ball land on the west side of the cannon? Since the Earth is rotating to the east at 1000 mph, this is what I would expect to happen. If the ball doesn't land on the west side of the cannon, what would that mean?
C Hammond both cannon and cannonball are travelling at 1000mph so the cannonball would land back inside the cannon I would have thought.
Alastair Douglas Yes,if ya shot one hundred cannon balls I'm sure at least a couple would come back down to dead center. But that would disprove the Coriolis Effect. Would it not?
+C Hammond I found this interesting and decided to do the calculation. For a simplified example (neglecting air resistance and assuming constant gravitational acceleration) it turns out that a cannon ball fired straight up with a velocity of 1000 feet per second would land about 94 feet to the west of where it was fired.
Ryan DeVos So, do you think that the 94 feet would be the actual result?
Yes, you are correct, it will land to the west. I have shot a bullet into the air and you can hear it land to the west.
Whether you toss into the Northern or Southern hemisphere the package will drift East (to the right).
In the Northern Hemisphere, drift will be to the right of initial motion. In the Southern Hemisphere, drift will be to the left of initial motion. Using "east" and "west" is probably not the best way to express this. For example, in the northern hemisphere, if an object is traveling south, it will get veered to the right (with would be east). But, if it is traveling north, it is getting veered to the right, but in this case, that would be west.
@@kenyanow8274 Perhaps I'm confused but, does not the given argument show that if you stand on the equator and pitch an object towards either pole in the direction of a longitude line then it will land to the East of that longitude line.
Peter Higgins, yes, but what if you're in the southern hemisphere at around 60 degrees south latitude and heave the object toward the equator? It will get veered to the left of initial motion (in this case, West).
how to solved drive gain problem(drive gain showing 1.2%).need drive gain 3%.
This DUDEEEEE 5:37
Thanks for the video.
This guy doesn't understand what the Coriolis effect is. He is confusing it with difference in velocities due to latitude.
TheRumpus agreed
TheRumpus I was looking to see if people really got into what he said. coz he is not true.
See www.nationalgeographic.org/encyclopedia/coriolis-effect/
So if guns and cannons have to account for the "coriolis effect" then why does the earth traveling through space at 67,000 mph not effect the bullet and only the 1000 mph rotation of the earth does?
Because the entire Earth is moving at the same 67,000 mph, so both the gun/cannon and its target are moving at that same speed, and therefore the bullet/cannonball has that velocity plus the velocity from being launched from the gun/cannon. Conversely, different points of the Earth are moving at different speeds due to rotation, so the gun/cannon and its target are moving at slightly different speeds from this, thereby requiring adjustment.
Because all of Earth's orbital motion is caused by gravity, and the Earth is in free fall around the sun. Ignoring the sun's tidal forces, there is no net constraint force necessary to keep you in the same position on Earth, against the sun's gravity, to keep you orbiting with the Earth. The sun's gravity is entirely "used up" in causing this motion, and thus both the orbital acceleration of Earth and the sun's gravity of Earth, will nullify each other in the reference frame of the Earth.
Also, that 67,000 mph orbital speed may seem high, but relative to the radius of curvature of the Earth's orbit, it is a hell of a lot less than the angular speed of the Earth's rotation. Angular speed matters more for the Coriolis effect than tangential speed.
Cool explaination but thats all it is.. Observe it in reality using the scientific method!
OK
Search for Foucault's pendulum. It's entirely based on that.
So if a plane travels from West to East it travels in the same direction of the rotation of the globe whereas if the plane travels from East to West it travels against the direction of the rotation of the globe. 🤔
Let me enlighten you: Earth is not a rotating globe guys!
Pilots do not take coriolis effect into account...
Well xplained.
👌
So, basically, the rate of change of the hypothetical object's velocity is part of a function of the cosine of the distance from the equator, as the equator is the origin for the Coriolis effect..
Okay. Nice explanation!
the wording was a bit weird to start. "The circumference of the Eart at [x]," never reaches zero, but a horizontal slice perpendicular to the axis of rotation could reach zero.Try not to confuse more people than you need to
Great
Not convinced . .
5:10- sorry, no. That’s not coriolis. Coriolis is simply an APPARENT DEFLECTION.
Sorry, yes.
Go back to your safe space.
0:23-aaaaaand you KNOW this how? Please cite the MAN who physically measured this. I’ll wait.
2:37-wtf?! Tell me: is there a difference between what a person physically feels closer to the axis on a merry-go-round than a person farther out?
The answer is yes.
That isn't Coriolis, Travis... that's centripetal acceleration you are feeling.
Good job asking a question that is not quite relevant to the topic.
@Flat Earth Data no.
Can you demonstrate it for yourself? Go put a pendulum on a lazy susan with the bob anywhere but the middle. Spin it.
Why does the bob deflect to the outside?
Answer: because it's accelerating INWARDS towards the centre.
@Flat Earth Data no...a lazy susan isn't the Earth, but it can be used to demonstrate centrpetal acceleration.
Didn't you do this in grade 11 or 12 in physics like the rest of Canada?
@Flat Earth Data Earth doesn't need to be a perfect circle. Neither does the lazy susan. The path the pendulum will make is a perfect circle, assuming proper construction of the lazy susan.
I don't need to find the COM in this example, but if the massive bob is a sphere of uniform material, it's quite easy. Do you need me to explain that?
1:43- calculations are not physical measurements. They are equations. How do you KNOW the equator is 24,000mi? Go.
This explanation is bogus. It has to do with viewing on a rotating frame of reference. There are many better videos about this on TH-cam.
I was looking for a mathematics explanation of the Coriolis Effect, like solving Newton’s 2nd law of motion in a rotating frame of reference, possible using Lagrange equation to make it easier, and I found this crap....this video is just about relative motion and even so: Totally wrong! 😢
I’m curious, Erik. What do you suppose is incorrect? I’ll also refer you to the following site for a similar explanation: www.nationalgeographic.org/encyclopedia/coriolis-effect/
If you’re looking for a detail derivation of the physics, this video is not for that (and clearly not for you). It is a concept based introduction to the topic.
Nosebleed!! Jowk mwahahahahaha
This is awful. I'm a musician but am aware of how wrong this is. If you put two pins on a globe in the two locations he mentions, perfectly in line north to south, and tie a string to them, and then spin the globe the string always stays straight. Yes, earth spins faster at equator but it's traveling a greater distance! So the two points remain in line regardless. This is silly. According to his logic a helicopter could fly perfectly north from the equator and end up hundreds of miles east and the passengers could literally watch the world spinning under them. Nonsense.
We're not talking about stationary objects pinned to the Earth. We are talking about moving objects not tied to the ground.
chuckhough, we are not talking about stationary objects pinned to the Earth. We are talking about moving objects not tied to the ground.
Eric Dubay 200 proofs explains it best
-_-
😭😭😵😵😵😳😳😬😬
Explain plane landing on a 1000 m/ hr spinning ball
This has already been explained multiple times here as well as elsewhere, which makes me doubt you actually want an answer and suspect that, more likely, you just want to insist that the Earth can't be round and spinning. It's not a case of a plane going ~150mph (approximate landing speed) landing on a surface spinning at 1,000mph, it's a case of a plane moving at ~1,150mph landing on a surface spinning at 1,000mph. This is because the plane always has the initial velocity of the spinning Earth *in addition to* its own velocity.
the pane & air is spinning along with the ball... just like how you would be able to maneuver and hover a drone inside a plane that you know is traveling 600 m/hr (traveling at a consistent speed, like the earth is rotating AND ALSO flying through space making a big circle around the sun) speeding up or slowing down makes g force
Don't fuck ears
BWHAHAHAHAHAAAAA
7:55-if you still believe this garbage at this point
No twist? They both twist, spin, rotate. Ffs.
So how come your beloved FE claims New Zealanders can see the sun thousands of miles further away that we can in the north?
Inside the Earth's environment the rotation of Earth has no effect.
That makes your explanation completely void.
I’m not sure what you’re talking about. Inside the Earth’s environment? You mean anywhere from the top of the atmosphere to sea surface? Here is another resource that discusses how Coriolis changes with elevation. geography.name/how-does-the-coriolis-effect-influence-wind-direction-at-different-heights/
@@kenyanow8274 I am glad you replied.
Your explanation is comparing the Earth to a Merry-go-round, isn't it?
But unlike the Merry-go-round if we make a jump from the Earth's surface we will not land at a different place viz. the Earth doesn't pass under our feet.
Because the Earth is a stationary for everything that is inside its atmosphere.
Plz clarify...
Please, sir.
@@binod3386 That might be a different video :). No, I did not refer to a merry-go-round. But, a merry-go-round analogy is not a bad one. In the analogy, no one is jumping off the merry-go-round. The Coriolis is the result of objects moving within a rotating reference frame.
Incorrect. We live on a Flat stationary plane.
Planes fly, silly
There is NO EARTH BASED CORIOLIS EFFECT
Your explanation makes no sense! The rate of rotation around the axis of the earth is constant no matter where on the surface u are standing. If Ecuador is due South of Nova Scotia, and u have a zip line that follows the trajectory of ur sling shot package, and the package is attached to the zip line, the package would end up in Nova Scotia after an hour. Your suggesting that if I jump at the equator, I will be slammed into the wall that is opposite the direction of the earth’s rotation, and if i do so in Nova Scotia then the distance to the wall that i would slam into could be half as far from the point i jump from, cause i’m moving slower? No wonder american kids suck at science and math! If their teachers can’t figure it out! Gravity is pulling everything down at a constant, that doesn’t change just because an item gets launched into the air... and the speed someones traveling at the equator would appear the exact same as someones speed standing at 60 degrees North Latitude relative to an observer in space because the person at 60 degrees latitude would appear on one horizon and disappear on the other horizon at the same times as the person standing at the equator, as a 2 dimensional view, they are moving at the same speed, but your suggesting that somehow the space observer can comprehend the depth distinction due to curvature, and i don’t think any one can easily distinguish the difference 2 dimensionally from observing in 3 dimensions of a sphere.
It makes sense. You are lacking comprehension.
5:32-wtf?! “Apparently of initial motion????” Is the classroom empty? Is no one actually listening to this mong???
6:05-“If everything is left undone???” Who taught you syntax and grammar??
No one accounts for the rotation of the earth when flying. If they did, then that would mean the earth is MOVING under the package.
How does this numpty change the direct line from Quito to NS to an offset of 500 miles? They were in line before and now they’re offset? FFS. HEAD toward NS and you’ll get there. FFS.
Coriolis effect only applies to rotating bodies. Doesn't apply to earth, that's why we never see any apparent drift.
Very bad vidio hai this kar
What a bunch of garbage. The spin of the earth changes the trajectory of an object moving directly north apparently 45°? How then do we explain an airplane flight from LA to New York taking more time than from New York to LA nonstop? If Coriolis is real then we should be able to go straight up in a helicopter, hover for a few hours and come straight back down and land in a completely different area but that does not happen.