Not only content is accurate and clear. He did go the extra length to play 4 roles .. The professor,the geek, the bored, and the intense scribe. Very nice ! Subbed and liked.
The production value on this deserves way more than the 874 views it has at the time of commenting. This has been very useful for a model I'm making for an engineering project. Thank you, good sir.
Did this some 10 years ago in a Physics 1 course to students, saw this video while I was trying to remind myself the method. Thanks for the erffort and good job. It is a pitty that the new textbooks are not covering the subject of num. analysis...
It is a very useful, if not slightly arcane, way of solving problems. It has a lot of uses and I think there is a lot to be learned from it. I agree with your assessment of the new textbooks.
Flipping Physics by the way i agree with the objections down at the comments... dealing with v square term in fluid friction is tricky, it cannot be simply seperated to vx^2 and vy^2 ... I think it should be something like shown in the following video: th-cam.com/video/uq4cg0MtzVo/w-d-xo.html Still can be handled by numerical analysis but one has to solve x and y simultaneously, they are not seperable...
I’m currently in 12th and am taking physics, I just wanna give you a huge amount of thanks. My teacher, while he is a good teacher, isn’t the best teacher for me. Your videos have made understanding this so much easier.
@@FlippingPhysics me too! I really enjoyed physics the way you present it, and I really just absorbed the information because of you! I can’t thank you enough.
Thumbs up for showing the calculations. Personally, I would use time increments of 0.001 seconds and Runge-Kutta to do the integration but this video makes it simple. I also like the fact the calculations were done using metric units. Edit, one more thing for the fanatics like me. The coefficient of drag, Cd, is not a constant. It is much higher than 0.5 when the velocities are low. So, when I calculate the coefficient of drag I use the Reynold number and index in the chart of the drag coefficient vs the Reynold number. I don't think it is right to use the vertical speed and the horizontal speed to calculate two different coefficient of drag. I wish there was a simple formula for calculated the coefficient of drag as a function of speed. But no, I have to make a cubic spline because the coefficient of drag as a function of the Reynold number is very non-linear or just plain weird.
Wow! This is by far the best explanation out there, You are my Hero! PS: It would be useful for me if you could show a video using the numerical method in MATLAB
Glad to be your hero. Sadly, I think I am going to let you down when I tell you I have no plans to create "a video using the numerical method in MATLAB". Sorry.
The formula used to update the position is not actually the Euler method. The Euler method would be xf = xi + vix*dt. The video uses what we would call the trapezoidal method xf = xi + vix*dt + 0.5*ax*dt^2. This is equivalent to using the average velocity during the interval dt to update the position instead of the initial value. Fortunately, the trapezodial method is a better method than the Euler method, so this “mistake” only makes the spreadsheet program more accurate.
This is great. I love that I keep learning through making these videos. To be clear, am I correct that I used the Euler method to determine velocities, however, I used the trapezoidal method to determine the positions?
When the acceleration is constant the right thing to do is always to have vxf = vxi + ax*dt. The Euler method and the trapezoidal method give the same velocity update. The only difference between the two methods in this case (constant acceleration) is the position update.
thank you for mentioning that. The video really helps me understand qudratic drag but it also made me confuse because the equation was diffrent from the Euler Method I already knew. I was doubting myself
@@FlippingPhysics in your video the ball was falling. I thought drag would always be an opposing force irrespective of up or down so would appear in the sheet as negative (just like for x direction). Though for gravitational force it's opposing in the upward direction but assisting when falling? These are just my guesses as I try to get my head around this. Great video though - far better than any others I have found on youtube and is what is encouraging me to try to understand this.
@@garethwilliams9695 In this example the ball is moving down, or in the negative y-direction. The drag force is opposite the direction the ball is moving, therefore, the drag force is in the positive y-direction. Glad you find the video helpful!
Greetings! Look, I'm developing a numerical methods course on matlab and I'd love to use your video on it, (if you sou allowe me). I have a question though, I took the excel sheet that you provide but as I increase the drag coefficient, I do see that the x displacement increases (which makes sense) but I do see that the velocity in the Y axies is bigger with drag that without it (i created columns for the position and velocity on the Y axies for both the drag and no drag case for each time t) and It seems as if when drag increases, the ball is able to reach higher altitudes than with no drag.... can I share the excel with you so you can guide me and see where I might be doing something wrong? Thanks
I think I found the problem with your excel sheet! Logically, Drag force is always going to have an opposite sign to the velocity in the X direction. Your sheet only considers tha case in which the ball is falling (with no initial velocity on the y axies). Adding *sign*(-1), solves this... please allow me to know if my hypothesis is correct! Happy new year btw :)
@@c0deattack710 Entering the mass as a negative value seems to "fix" this. Otherwise, the vertical acceleration calculation comes out positive, which it cannot be. It then agrees with two spreadsheets that I worked out on my own. However, there seems to be some other factor not accounted for. Trying to get that Lacrosse ball up to 1000 m height takes some huge launch speed like 7000 m/sec. Actual artillery shells go way higher, what are we missing? I'm guessing that somehow the drag is much less for some distance at the beginning of the trajectory, whereas the calculated velocity during the first 0.1 sec shows a 50% drop. Visualize that the air in direct line of gun must get blown foreward, and gases behind the exiting shell also have momentum.
+Shakila Taylor Break the initial velocity into its x and y components (Like I do in my Nerd-A-Pult video: www.flippingphysics.com/nerd-a-pult.html). Those become the initial velocities in the x and y directions in the spreadsheet.
Great video, however just to clear up an inquiry, you squared he value of initial velocity for the horizontal component (Vxi) to calculate the force of drag (Fdx) in that component however doesn't that mean that the horizontal and vertical components are no longer coupled? Shouldn't it be -Fdx=-kvvcos(theta) which equals -Fdx=-Kvivxi where vi=sqrt(vxi^2+vyi^2)?
If you decompose the Fdrag for instance in Fdrag x you would get to following, right? So vx^2 would be incorrect? Fdx=Fd ∙ cos(alpha). Fdx=0.5 ∙ 𝜌 ∙ v^2 ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑 ∙ cos(alpha) v ∙ cos(alpha)= vx Fdx=0.5 ∙ 𝜌 ∙ v ∙ vx ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑 v = √(𝑣𝑥^2 + 𝑣𝑦^2) 𝐹𝑑𝑥 = 0.5 ∙ 𝜌 ∙ √(𝑣𝑥^2 + 𝑣𝑦^2) ∙ 𝑣𝑥 ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑 Furthermore, I love your way of explaining.
Flipping Physics in actuality, you can somewhat do it. You need to solve each component’s differential equations. Just a heads up though, your y component becomes a piecewise function, so you would need to solve for the time for the max height and write an if statement in excel.
If you follow the link in the description: www.flippingphysics.com/the-euler-method.html On that page you will find a link labelled "The Excel file created in the video" Download that. The same Excel file can be used with an initial velocity at an angle, you have to input the initial components of the initial velocity.
@@FlippingPhysics hi sorry must be me but I'm not seeing it. I tried to rebuild your sheet but noted possibly that the vertical drag force or had to change direction on the "up" of the arc. Sadly the numbers I'm getting out of it don't match with another solution I had produced it's producing much larger distances. I.e. the 0deg to 5deg difference is alot more than I expected. Would appreciate more discussion to make sure I've got this setup right.
The drag coefficient, C, can be extremely variable, especially at low speeds and for small, smooth objects like glass marbles. Most times it's reasonable to use the theoretical values from the tables, but sometimes you'll need to measure the drag coefficient experimentally if you want to get accurate results.
You end up referencing a cell you didn't intend to reference. Suppose in cell B4, you enter the formula =A2*A4^2, where you intend A2 to reference a constant (suppose it is equal to 10), and A4 to reference a variable. You then populate cells A4 thru A24 with the numbers 1 thru 20. As you drag the formula you entered in Cell B4, all the way to B24, with the intent that you will reference A2 as a constant, and the corresponding A-cell as the variable you square, you will actually be referencing cells other than A2 to determine this constant. Drag cell B4 to cell B5, and the formula it will automatically generate, will be =B3*B5^2, which will return 0, since B3 is initially empty. Drag to cell B6, and it will now generate the formula =B4*B6^2, which will return 9. What we intended was for cell B4 to equal 10*1^2 = 10, cell B5 to equal 10*2^2 = 40, and cell B6 to equal 10*3^2 = 90. What we got instead was cell B4=10, cell B5=0, and cell B6 = 9. The proper way to do this, is to write =$A$2*A4^2 in cell B4, and then drag that down. Or, assign a custom variable name of K to cell A2, and write =K*A4^2 in cell B4.
Hey Great video cleared up a lot of stuff, Just wondering, a question very similar to this came up in an exam in which the question gave the Force of drag formula and asked you to find highest point, distance travelled etc, is there any way of finding Maxima shortcuts or would you have to bang out the results on your calculator the same way you've found it?
If wind (-head or +tail) was included on top of drag, then what, a change in velocity every step of the way? It's not as simple as increasing or deceasing the drag is it?
If the wind is parallel to the direction of the velocity, then it should be as simple as increasing or decreasing the drag. However, if the wind is not parallel to the direction of the velocity, then you should have another draw force in the direction of the wind.
Thanks. OK, so maybe I should qualify "simple." Let's say it's parallel, reduce or increase Fdrag or Dragx (N) by v^2 object +/-v wind? so Fdragx = -pv^2 +/- vwind DA/2m?
@@FlippingPhysics Thank you so much for this video. I have been looking for how to do this for years. Do you know how to add wind? I need to incorporate tail wind into my spreadsheet. I use it for arrows. 150 lb yew bow with a 95.9 gram arrow. 45 degree angle, 53.32 m/s as initial velocity, 43.3 m/s at the end of the trajectory. Range around 244 meters with a 9 m/s tail wind. Drag coefficient is 2.0. If the arrow is 53.6 gram the drag cof is 1.8. Surface area is 0.0001779768698 m^2. Position is 1.6 m over ground. Info on page 410 in the Great Warbow by Hardy and Strickland.
Interesting. My guess is what you are really asking is "Is the force of drag equation valid in liquids?". Fluid means a substance which does not have a fixed shape which means it can be a liquid or a gas. My answer, as far as I know is: The force of drag equation is valid in all fluids. Realize the drag force is just one of the forces acting on objects in liquids; the buoyant force is another force acting on an object in a liquid.
Thank you, i also noticed, there is great variation in the final velocity and position values when considering a time interval of 1/1000 instead for the same interval.
You need to use stoke's law for a sphere moving through a liquid. In this case, the drag is proportional to v, not v^2. Do an internet search for Stoke's law and you'll find plenty of info on this.
That's right. Stoke's Law. It has to be a small object and the flow needs to be laminar. Here are a couple of websites about it. Thanks for the suggestion! aerosol.ees.ufl.edu/aerosol_trans/section03_b.html scienceworld.wolfram.com/physics/StokesVelocity.html
Hi thank you very much for this. I'd like to know if I could do the same with an object with initial position in the y direction of 1meter, and this object should reach a final y position of 1.60 meter, and a x final position of 30 meters. I've the initial velocity, and it has an angle.
Flipping Physics Thanks, Just one more question. I saw the discussion in the comments below, and I didn't get it. Should I break de drag force into the x and y components?
I am 95% sure that the approach in the discussion below ends up being identical to mine. I split the velocity into components first and then determined the force of drag components from there. I am pretty sure that the other approach below breaks the force of drag into components. It makes more sense to me to break the velocity into components first.
Because this isn't as simple as a Calculus problem, but rather it is a differential equations problem. You have a lot of inter-dependent derivative terms, that make it incredibly difficult to solve analytically. There are analytic solutions, but they are far beyond the scope of a high school class, even with the background of Calculus. Complicated problems like this, often do exist with no analytical solution in closed-form, and the only way to get a solution that is good enough for engineering purposes, is to use a numerical computing method as featured in this video, for an approximate answer. This is what finite elements and computational fluid mechanics would do for you.
Hi guys, I have a question. If the aceleration is variable can I use Torriceli equation to find the position? S = S0 + v0Δt + aΔt² / 2 What I mean is the aceleration wil change in this Δt, but will we use the same aceleration for all Δt? Thanks!
If acceleration is not constant during the time interval Δt, then that particular equation is not valid. You have to establish a as a function of t, and use calculus to integrate it twice to get the displacement that happens over the time interval Δt. All of your standard 5 kinematic formulas rely on the premise that acceleration is constant. There are calculus versions of these formulas that enable you to account for a variable acceleration as a function of time. Sometimes you might know how acceleration varies as a function of time, other times you might know how acceleration varies as a function of distance (such as with a spring force). You might even know instead how acceleration varies with velocity, as is the case for air drag, where you'll need differential equations to solve for kinematic equations as functions of time.
Instead of starting with initial x and y velocities, start with a velocity magnitude and angle. Use trig ratios sin and cos to get the x and y components for your initial and work from there. At every time step, using arctan on the y and x components will give you the angle.
For a sphere, there is only one force from the air, which is the force of drag. You can get more air forces with an initial spin on the sphere, but not lift. Lift comes about when there is a shape with a preferred direction to move through the air, like a flat board. A flat board would "feather" the air with its speed parallel to the long direction. Drag acts opposite its velocity relative to the air., and lift acts perpendicular to its velocity relative to the air.
If alpha is the angle between the velocity vector and the horisontal then: cos(alpha)=V(x direction)/V and Sum(F(x))=-F(x)=-Fcos(alpha)=-FV(x)/V)=(-0.5pV^2DAV(x))/V=-0.5pDAV(x)V= -0.5pDAV(x)*sqrt(V(x)^2+V(y)^2))=mA(x direction) --> A(x direction)= (-pDAV(x)*sqrt(V(x)^2+V(y)^2))/(2m) which is not quite the same...
Dude, it's really difficult to convey physics equations in TH-cam comments. Sorry, however, I am not even going to try to parse together what you are saying here.
I have also spotted the mystake - the point is you cannot split simply F drag into vx and vy component as presented in the video and corrected by @Fredde
Flipping Physics the formula you have written says that F(drag)=1/2pV^2DA. This force is in the opposite direction of the velocity V of the projectile. You split this force up in components the same way you do with the velocity for example: v(x direction)=v*cos(angle with horisontal) and so F(drag, x direction)= -F*cos(angle with horisontal) . If you do this you will end up with a different formula. In the dragforce formula you have v squared. When you multiply v squared by cos(angle) you will end up with v(x-direction)*v and not v(x-direction) squared
Try implementing the spreadsheet and see what you get. Then try the same experiment with balls you can find in reality, and try the experiment for yourself. What you should expect is that the greater the mass-to-R^2 ratio, the closer its motion will be to a free projectile analysis that neglected air drag. So I would expect a bowling ball to be more like a free projectile than a volleyball, but I would also expect a billiard ball to be more like a free projectile than a bowling ball. Whereas an inflatable beach ball should have the most influence of drag.
*Value for ball size* : National Operating Committee on Standards for Athletic Equipment *Value for Cd of ball* : National Aeronautics and Space Administration *Value for Density of air* : aBoUt.cOm
Not only content is accurate and clear. He did go the extra length to play 4 roles .. The professor,the geek, the bored, and the intense scribe. Very nice ! Subbed and liked.
Actually, I have 3 identical brothers who agree to play the roles. But I am glad you appreciate this!
@@FlippingPhysics 🤔
The production value on this deserves way more than the 874 views it has at the time of commenting. This has been very useful for a model I'm making for an engineering project. Thank you, good sir.
You are welcome and thank you for your kind words. I wish the video had more views as well. 😊
Did this some 10 years ago in a Physics 1 course to students, saw this video while I was trying to remind myself the method. Thanks for the erffort and good job. It is a pitty that the new textbooks are not covering the subject of num. analysis...
It is a very useful, if not slightly arcane, way of solving problems. It has a lot of uses and I think there is a lot to be learned from it. I agree with your assessment of the new textbooks.
Flipping Physics by the way i agree with the objections down at the comments... dealing with v square term in fluid friction is tricky, it cannot be simply seperated to vx^2 and vy^2 ...
I think it should be something like shown in the following video:
th-cam.com/video/uq4cg0MtzVo/w-d-xo.html
Still can be handled by numerical analysis but one has to solve x and y simultaneously, they are not seperable...
6 years later but we are covering the topic for python, but not really in our textbooks!
I’m currently in 12th and am taking physics, I just wanna give you a huge amount of thanks. My teacher, while he is a good teacher, isn’t the best teacher for me. Your videos have made understanding this so much easier.
I'm so glad you found my channel!
@@FlippingPhysics me too! I really enjoyed physics the way you present it, and I really just absorbed the information because of you! I can’t thank you enough.
This was an amazing video, very educational and funny at the same time
Thank you!!
Im making a big projekt in school, which is about air this subject. You completely saved my project! Thank you so much. Lots of love from Denmark
Love it! You are absolutely welcome. Glad to know I can help you learn! I hope the project works out well for you.
Dude, I've been living my life using Excel without absolute references, just hating it when I need to paste formulas... what a game changer!!
So glad to help!
What an absolutely outstanding video...
Thanks!
Thumbs up for showing the calculations. Personally, I would use time increments of 0.001 seconds and Runge-Kutta to do the integration but this video makes it simple. I also like the fact the calculations were done using metric units. Edit, one more thing for the fanatics like me. The coefficient of drag, Cd, is not a constant. It is much higher than 0.5 when the velocities are low. So, when I calculate the coefficient of drag I use the Reynold number and index in the chart of the drag coefficient vs the Reynold number. I don't think it is right to use the vertical speed and the horizontal speed to calculate two different coefficient of drag.
I wish there was a simple formula for calculated the coefficient of drag as a function of speed. But no, I have to make a cubic spline because the coefficient of drag as a function of the Reynold number is very non-linear or just plain weird.
Wow!
This is by far the best explanation out there, You are my Hero!
PS: It would be useful for me if you could show a video using the numerical method in MATLAB
Glad to be your hero. Sadly, I think I am going to let you down when I tell you I have no plans to create "a video using the numerical method in MATLAB". Sorry.
haha, Not to worry
this was such a cool video. thanks mr. p!
This is amazing!!! You deserve all the physics subs on youtube!
Thanks. A lot went in to making this video.
The explanation is helpfull to understand numerical modelling
thank you
You are welcome.
The formula used to update the position is not actually the Euler method. The Euler method would be xf = xi + vix*dt. The video uses what we would call the trapezoidal method xf = xi + vix*dt + 0.5*ax*dt^2. This is equivalent to using the average velocity during the interval dt to update the position instead of the initial value. Fortunately, the trapezodial method is a better method than the Euler method, so this “mistake” only makes the spreadsheet program more accurate.
This is great. I love that I keep learning through making these videos. To be clear, am I correct that I used the Euler method to determine velocities, however, I used the trapezoidal method to determine the positions?
When the acceleration is constant the right thing to do is always to have vxf = vxi + ax*dt. The Euler method and the trapezoidal method give the same velocity update. The only difference between the two methods in this case (constant acceleration) is the position update.
thank you for mentioning that. The video really helps me understand qudratic drag but it also made me confuse because the equation was diffrent from the Euler Method I already knew. I was doubting myself
Shouldn't the drag in the y direction be negative like in x direction? drag is always an opposing force?
Is the object moving up or down?
@@FlippingPhysics in your video the ball was falling. I thought drag would always be an opposing force irrespective of up or down so would appear in the sheet as negative (just like for x direction). Though for gravitational force it's opposing in the upward direction but assisting when falling? These are just my guesses as I try to get my head around this. Great video though - far better than any others I have found on youtube and is what is encouraging me to try to understand this.
@@garethwilliams9695 In this example the ball is moving down, or in the negative y-direction. The drag force is opposite the direction the ball is moving, therefore, the drag force is in the positive y-direction. Glad you find the video helpful!
@@FlippingPhysics Oh I see! haha. Thanks that's made it clear and I had my understanding the wrong way around. Thank you.
@@garethwilliams9695 No worries. This is what learning is all about.
Greetings!
Look, I'm developing a numerical methods course on matlab and I'd love to use your video on it, (if you sou allowe me). I have a question though, I took the excel sheet that you provide but as I increase the drag coefficient, I do see that the x displacement increases (which makes sense) but I do see that the velocity in the Y axies is bigger with drag that without it (i created columns for the position and velocity on the Y axies for both the drag and no drag case for each time t) and It seems as if when drag increases, the ball is able to reach higher altitudes than with no drag.... can I share the excel with you so you can guide me and see where I might be doing something wrong?
Thanks
I think I found the problem with your excel sheet! Logically, Drag force is always going to have an opposite sign to the velocity in the X direction. Your sheet only considers tha case in which the ball is falling (with no initial velocity on the y axies).
Adding *sign*(-1), solves this... please allow me to know if my hypothesis is correct!
Happy new year btw :)
@@c0deattack710 Entering the mass as a negative value seems to "fix" this. Otherwise, the vertical acceleration calculation comes out positive, which it cannot be. It then agrees with two spreadsheets that I worked out on my own.
However, there seems to be some other factor not accounted for. Trying to get that Lacrosse ball up to 1000 m height takes some huge launch speed like 7000 m/sec. Actual artillery shells go way higher, what are we missing? I'm guessing that somehow the drag is much less for some distance at the beginning of the trajectory, whereas the calculated velocity during the first 0.1 sec shows a 50% drop.
Visualize that the air in direct line of gun must get blown foreward, and gases behind the exiting shell also have momentum.
So what do you do if the object is launched at an angle?
+Shakila Taylor Break the initial velocity into its x and y components (Like I do in my Nerd-A-Pult video: www.flippingphysics.com/nerd-a-pult.html). Those become the initial velocities in the x and y directions in the spreadsheet.
At 18:26 shouldn't the acceleration in the final Y position equation be the acceleration in the y dimension not x?
Bummer. Yes, that is a typo. That should be the acceleration in the y direction in that equation.
Lovely, great vid, thank you very much.
You are welcome!
Great video, however just to clear up an inquiry, you squared he value of initial velocity for the horizontal component (Vxi) to calculate the force of drag (Fdx) in that component however doesn't that mean that the horizontal and vertical components are no longer coupled? Shouldn't it be -Fdx=-kvvcos(theta) which equals -Fdx=-Kvivxi where vi=sqrt(vxi^2+vyi^2)?
You are my saviour
Glad to help.
Since the force of drag is always changing, how can you determine the amount of energy lost due to drag? Please reply asap
I think you can calculate the power using P=F*v= -kv³
And then you use the Euler method to calculate the work at any instant of time, using
W=P*∆t
If you decompose the Fdrag for instance in Fdrag x you would get to following, right? So vx^2 would be incorrect?
Fdx=Fd ∙ cos(alpha).
Fdx=0.5 ∙ 𝜌 ∙ v^2 ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑 ∙ cos(alpha)
v ∙ cos(alpha)= vx
Fdx=0.5 ∙ 𝜌 ∙ v ∙ vx ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑
v = √(𝑣𝑥^2 + 𝑣𝑦^2)
𝐹𝑑𝑥 = 0.5 ∙ 𝜌 ∙ √(𝑣𝑥^2 + 𝑣𝑦^2) ∙ 𝑣𝑥 ∙ 𝐴𝐵𝑎𝑙 ∙ 𝐶𝑑
Furthermore, I love your way of explaining.
is there a way to solve this exact thing but using integration?
+David Rodriguez Nope. I don't believe so.
Flipping Physics in actuality, you can somewhat do it. You need to solve each component’s differential equations. Just a heads up though, your y component becomes a piecewise function, so you would need to solve for the time for the max height and write an if statement in excel.
Thank you, I found the content very helpful and understandable:)
can you solve the differential equation for this and get the same answer?
Thank you mr p . I can say that I'm a happy physics student : ) : ) .
+Chary Nud I am glad I can contribute at least one more happy physics student to the world!
hi i tried to download the excel but struggled to find it. also do you have this for a trajectory approach where the item starts at a launch angle?
If you follow the link in the description:
www.flippingphysics.com/the-euler-method.html
On that page you will find a link labelled "The Excel file created in the video"
Download that.
The same Excel file can be used with an initial velocity at an angle, you have to input the initial components of the initial velocity.
@@FlippingPhysics hi sorry must be me but I'm not seeing it.
I tried to rebuild your sheet but noted possibly that the vertical drag force or had to change direction on the "up" of the arc. Sadly the numbers I'm getting out of it don't match with another solution I had produced it's producing much larger distances. I.e. the 0deg to 5deg difference is alot more than I expected. Would appreciate more discussion to make sure I've got this setup right.
The drag coefficient, C, can be extremely variable, especially at low speeds and for small, smooth objects like glass marbles. Most times it's reasonable to use the theoretical values from the tables, but sometimes you'll need to measure the drag coefficient experimentally if you want to get accurate results.
how would I implement angle?
What happens to your spreadsheet if you don't use absolute references at the relevant points?
You end up referencing a cell you didn't intend to reference. Suppose in cell B4, you enter the formula =A2*A4^2, where you intend A2 to reference a constant (suppose it is equal to 10), and A4 to reference a variable. You then populate cells A4 thru A24 with the numbers 1 thru 20. As you drag the formula you entered in Cell B4, all the way to B24, with the intent that you will reference A2 as a constant, and the corresponding A-cell as the variable you square, you will actually be referencing cells other than A2 to determine this constant. Drag cell B4 to cell B5, and the formula it will automatically generate, will be =B3*B5^2, which will return 0, since B3 is initially empty. Drag to cell B6, and it will now generate the formula =B4*B6^2, which will return 9.
What we intended was for cell B4 to equal 10*1^2 = 10, cell B5 to equal 10*2^2 = 40, and cell B6 to equal 10*3^2 = 90. What we got instead was cell B4=10, cell B5=0, and cell B6 = 9. The proper way to do this, is to write =$A$2*A4^2 in cell B4, and then drag that down. Or, assign a custom variable name of K to cell A2, and write =K*A4^2 in cell B4.
Hey Great video cleared up a lot of stuff, Just wondering, a question very similar to this came up in an exam in which the question gave the Force of drag formula and asked you to find highest point, distance travelled etc, is there any way of finding Maxima shortcuts or would you have to bang out the results on your calculator the same way you've found it?
+MrICEDBlack I don't know of any Maxima / Minima shortcuts. You have to "bang out the results (in a spreadsheet) the same way (I) found it".
If wind (-head or +tail) was included on top of drag, then what, a change in velocity every step of the way? It's not as simple as increasing or deceasing the drag is it?
If the wind is parallel to the direction of the velocity, then it should be as simple as increasing or decreasing the drag. However, if the wind is not parallel to the direction of the velocity, then you should have another draw force in the direction of the wind.
Thanks. OK, so maybe I should qualify "simple." Let's say it's parallel, reduce or increase Fdrag or Dragx (N) by v^2 object +/-v wind? so Fdragx = -pv^2 +/- vwind DA/2m?
@@FlippingPhysics Thank you so much for this video. I have been looking for how to do this for years. Do you know how to add wind? I need to incorporate tail wind into my spreadsheet. I use it for arrows. 150 lb yew bow with a 95.9 gram arrow. 45 degree angle, 53.32 m/s as initial velocity, 43.3 m/s at the end of the trajectory. Range around 244 meters with a 9 m/s tail wind. Drag coefficient is 2.0. If the arrow is 53.6 gram the drag cof is 1.8. Surface area is 0.0001779768698 m^2. Position is 1.6 m over ground. Info on page 410 in the Great Warbow by Hardy and Strickland.
Dude, You're awesome, i only passed my IB Extended Essay because of let me know if you want to see it.
Congrats on passing IB. I have heard it is quite a challenge.Sure, I would enjoy taking a look at it. jon at flippingphysics dot com
The link goes to a "Not Found - Error 404" page.
hi im in Ib as well and this is for my Math IA
Great video!!! :)
Is it fine to use this for motion of a ball in fluids, considering viscous drag and buoyant force
Interesting. My guess is what you are really asking is "Is the force of drag equation valid in liquids?". Fluid means a substance which does not have a fixed shape which means it can be a liquid or a gas. My answer, as far as I know is: The force of drag equation is valid in all fluids. Realize the drag force is just one of the forces acting on objects in liquids; the buoyant force is another force acting on an object in a liquid.
Thank you, i also noticed, there is great variation in the final velocity and position values when considering a time interval of 1/1000 instead for the same interval.
You need to use stoke's law for a sphere moving through a liquid. In this case, the drag is proportional to v, not v^2. Do an internet search for Stoke's law and you'll find plenty of info on this.
That's right. Stoke's Law. It has to be a small object and the flow needs to be laminar. Here are a couple of websites about it. Thanks for the suggestion!
aerosol.ees.ufl.edu/aerosol_trans/section03_b.html scienceworld.wolfram.com/physics/StokesVelocity.html
Hi there, what if i dont know the initial horizontal velocity of my projectile?? Yours was 10m/s...
what about using iteration on excell
Hi thank you very much for this. I'd like to know if I could do the same with an object with initial position in the y direction of 1meter, and this object should reach a final y position of 1.60 meter, and a x final position of 30 meters. I've the initial velocity, and it has an angle.
+Luis Felipe Sanches I bet you could work it out. The excel file is posted on my website, if that helps.
Flipping Physics Thanks, Just one more question. I saw the discussion in the comments below, and I didn't get it. Should I break de drag force into the x and y components?
I am 95% sure that the approach in the discussion below ends up being identical to mine. I split the velocity into components first and then determined the force of drag components from there. I am pretty sure that the other approach below breaks the force of drag into components. It makes more sense to me to break the velocity into components first.
Flipping Physics You're awesome man, thank you.
You are welcome and thanks. 😀
why was calculus not involved in the calculations?
Because this isn't as simple as a Calculus problem, but rather it is a differential equations problem. You have a lot of inter-dependent derivative terms, that make it incredibly difficult to solve analytically. There are analytic solutions, but they are far beyond the scope of a high school class, even with the background of Calculus.
Complicated problems like this, often do exist with no analytical solution in closed-form, and the only way to get a solution that is good enough for engineering purposes, is to use a numerical computing method as featured in this video, for an approximate answer. This is what finite elements and computational fluid mechanics would do for you.
THANK YOU SO MUCH :)
Hi guys, I have a question.
If the aceleration is variable can I use Torriceli equation to find the position?
S = S0 + v0Δt + aΔt² / 2
What I mean is the aceleration wil change in this Δt, but will we use the same aceleration for all Δt?
Thanks!
If acceleration is not constant during the time interval Δt, then that particular equation is not valid. You have to establish a as a function of t, and use calculus to integrate it twice to get the displacement that happens over the time interval Δt. All of your standard 5 kinematic formulas rely on the premise that acceleration is constant.
There are calculus versions of these formulas that enable you to account for a variable acceleration as a function of time. Sometimes you might know how acceleration varies as a function of time, other times you might know how acceleration varies as a function of distance (such as with a spring force). You might even know instead how acceleration varies with velocity, as is the case for air drag, where you'll need differential equations to solve for kinematic equations as functions of time.
Can anyone help me explain how to apply the angle of throw into this?
Instead of starting with initial x and y velocities, start with a velocity magnitude and angle. Use trig ratios sin and cos to get the x and y components for your initial and work from there. At every time step, using arctan on the y and x components will give you the angle.
does this include lift?
For a sphere, there is only one force from the air, which is the force of drag. You can get more air forces with an initial spin on the sphere, but not lift.
Lift comes about when there is a shape with a preferred direction to move through the air, like a flat board. A flat board would "feather" the air with its speed parallel to the long direction. Drag acts opposite its velocity relative to the air., and lift acts perpendicular to its velocity relative to the air.
from where did he get the velocity
This is an extension of a previous video where the velocity of the car is shown.
www.flippingphysics.com/projectile-motion-problem-part-1-of-2.html
If alpha is the angle between the velocity vector and the horisontal then: cos(alpha)=V(x direction)/V and
Sum(F(x))=-F(x)=-Fcos(alpha)=-FV(x)/V)=(-0.5pV^2DAV(x))/V=-0.5pDAV(x)V=
-0.5pDAV(x)*sqrt(V(x)^2+V(y)^2))=mA(x direction)
--> A(x direction)= (-pDAV(x)*sqrt(V(x)^2+V(y)^2))/(2m) which is not quite the same...
Dude, it's really difficult to convey physics equations in TH-cam comments. Sorry, however, I am not even going to try to parse together what you are saying here.
It indeed looks kinda hairy
I have also spotted the mystake - the point is you cannot split simply F drag into vx and vy component as presented in the video and corrected by @Fredde
Interesting. What is your argument as to why it is not correct to break the force of drag into its x and y components?
Flipping Physics the formula you have written says that F(drag)=1/2pV^2DA. This force is in the opposite direction of the velocity V of the projectile. You split this force up in components the same way you do with the velocity for example: v(x direction)=v*cos(angle with horisontal) and so F(drag, x direction)= -F*cos(angle with horisontal) . If you do this you will end up with a different formula. In the dragforce formula you have v squared. When you multiply v squared by cos(angle) you will end up with v(x-direction)*v and not v(x-direction) squared
Isn't it 1.225 instead of 1.275?
In this experiment, you're using ball with a small size and mass. What if the ball is bigger than that? Let say that ball has 51.17kg and radius 1m
Try implementing the spreadsheet and see what you get. Then try the same experiment with balls you can find in reality, and try the experiment for yourself.
What you should expect is that the greater the mass-to-R^2 ratio, the closer its motion will be to a free projectile analysis that neglected air drag. So I would expect a bowling ball to be more like a free projectile than a volleyball, but I would also expect a billiard ball to be more like a free projectile than a bowling ball. Whereas an inflatable beach ball should have the most influence of drag.
*Value for ball size* : National Operating Committee on Standards for Athletic Equipment
*Value for Cd of ball* : National Aeronautics and Space Administration
*Value for Density of air* : aBoUt.cOm
my teacher linked this video to a joke problem he assigned us and i skipped to the middle and was terrified
Flipping Physics:
"...I converted from English to metric units..."
The English physicist:
~cries into tea.~
You mean the English physicists like James Prescott Joule and Isaac Newton?
I have been trying to do this on my own for 3 days now 😂
刘思佳 Dude. I'm sorry, however, I can't solve a problem you've been assigned by a teacher. It's some sort of teacher code or something.
This is like Chinese to me .. Was expecting a video that will explain it to me like I was a five year old child
Wow, can we just create an electronic sensor or use a radar to measure the velocity and accelaration? using radars and lidars... LOL.
Anyone else here from the BYUI ME-142 class?