Thank you for this absolutely beautiful explanation. This video should be the first one that should come up when looking for 'GForce'! Great video..thank u
Sir Thank you very much for your simplified explanation. Appreciate your effort. Had one doubt, feeling lighter as lift comes down (11.03min) is understandable however while roller coaster comes down (13.24min) why we feel heavier? Is centripetal acceleration different from linear?
ನಾಗರಾಜ್ ರಾವ್ thanks for your comment. On the rollercoaster, it's when you're at the bottom of a dip that you feel heavier. That is, as your vertical velocity slows and you start to go back up again. Sorry if that wasn't so clear!
Thank you for your infomative vdo. Again please confirm whether my understanding is correct. If I would go above speed of sound 1,100 ft per sec from zero it would mean G force would be 1,100ft/9.8 m act on my body would be around 33 g?? But if I would travel continuously on that speed the inertial would fall to normal. Is my understanding correct?
Pretty close. You are correct that if you travel at a constant velocity, your g-force would be "normal". ie: just 1 g due to gravity. For your speed of sound scenario, a) you should work in consistent units; preferably SI units. Convert to ft/s to m/s. and b) the g-force on your body would depend on how quickly you accelerated, not your final speed. If you went from 0 to the speed of sound (343m/s) in, say, 10 seconds, that is an acceleration of 34.3 m/s each second, or 34.3 m/s/s, which creates a g-force of 34.3/9.8 = 3.5g. You could then vector add the 1g due to gravity to the 3.5g due to acceleration to determine the overall g-force. I hope that helps!
Hi, nice video but i think the fact that you mentioned that astronauts have zero gravity in space is not right. isn't it supposed to be much smaller than that on earth, but not zero?
Hi Irwan, thanks, I'm glad you liked the video. You are correct, even out in space there is gravity, just less and less the further out you go. Interestingly, the reason astronauts are 'weightless' is not because they're away from the Earth's gravity, but because they're in orbit. An object in orbit is actually in freefall, and an object in freefall is weightless. So I probably should have used that as my example. Have a great day!
Does this mean hypothetically if a lift accelerates downwards at 9,81 m/s2 we would be weightless like in a vacuum? Sorry for my english hope you understand.
An object at rest must be resisting the rotaional/revolutionary forces of earth or the fluids in the body must be resisting the atmospheric pressure due to gases?
Sorry if this is a dumb question but say you’re only accelerating horizontally, would you not have to + the acceleration due to gravity as it’s acting vertically downwards?
Correct. The total g force would be calculated by the vector addition of the downward and horizontal components, resulting in a 'diagonal" downward and backward g force.
Hi Timo - Any time you have an acceleration, you feel an "inertial force" against the acceleration (again, this is not really a force, but is your inertia resisting acceleration as per Newton's first law. So you can't draw it in a free body diagram. But it *feels* like a force. eg: when you accelerate forwards in a car you 'feel' an inertial 'force' pulling you backwards, even though in a FBD you'd only draw a force pushing you forward). Now in freefall, you are accelerating downwards, so by my argument above, there should be a felt inertial 'force' upward. But we also feel gravity pulling us downward which cancels out the 'feeling' of the upward inertial 'force'. I hope that helps. PS - The other way of analysing g-force is to consider the net force of the surrounds acting on you. That's easier mathematically, (especially for free-fall) but not quite as intuitive. If I get some time, I might create a video showing that approach also.
@@timo41 There isn't a force upwards (apart from air resistance which is neglected). There is an "apparent force" which is just the body's resistance to the acceleration caused by the downward force of gravity. G-Force is essentialy the apparent weight of the body. The body's inertia results in the the upward G-Force as the inertia is opposite to the downward acceleration.
If an object is falling, it will naturally accelerate at 9.8m/s/s=1g (or less due if air resistance becomes significant). Never more. However, if an object is pushed downward by some other force with an acceleration greater than 9.8m/s/s, then a negative g force is experienced. For example, if you're in an airplane and it hits turbulence and accelerates quickly downwards, if you don't have your seatbelt on, you may find yourself feeling lifted out of your seat!
So if you travel at ~9.8 m/s upwards, and gravity is pulling you down at ~9.8 m/s, you experience two Gs. But in free fall, you float and feel weightless. Aren't you still being pulled down at 9.8, therefore feel the weight?
Hi Ahmed - When you are in freefall, gravity is pulling you down at 1g AND you are accelerating downward at 1g. So, mathematically, the total G-Force you feel is +1g (due to gravity) and -1g (opposite to your downward acceleration). Which totals zero g. This matches with what you'd expect if you do a 'thought experiment'. If you were in a lift accelerating downward at 9.8m/s, you would "float" inside the lift, because both you and the lift are accelerating downward at the same rate. I hope I've understood your question correctly, and my answer helps!
Basically you feel zero gs, acceleration wise. You get pulled down by gravity and also pushed up by inertia. As you float down you feel zero g-force. As he previously mentioned, g-force is the weight you feel because of gravity or acceleration. You will feel zero g-force because of the upward and downward pull of both gravity and inertia. Anything else you feel by free falling is probably because of air resistance which is a completely different topic. You must be familiar with the ball and feather gravity experiment, right?
God math has never been my strong suit. Ok, the rocker in space is actually relevant to the topic I'm interested in. I'll explain the issue. I play Star citizen a space sim game. They've implemented a G force mechanism that causes you to black out. I argue this is unrealistic. Most ship speeds in the game are in meters like 1000 meters. I'm unaware of whether it's meters per second, meters per second squared, or just meters and hour. (I suspect it's meters per second.) Seeing as most people black out at 4-5Gs, I figured at 50 m/s squared you're blacking out. However, at 1000 m/s squared, you'd be liquefied. However, this is meters per second squared. Could you give examples for other units of measuring speed like mph and meters per second?
Hi @kazzdevlin5339. Its not the speed (m/s) that determines the g-force you feel, its the acceleration (m/s/s or metres per second squared). This is, how quickly your speed is changing. If you went from 0m/s (speed) to 100 m/s (also a speed) in, say, 3 seconds, that means your speed is changing at 33.3 metres per second every second. In other words, your acceleration is 33.3 metres per second per second. So that woule be a bit over 3 "g"s of accleration. (33.3/9.8). If you went from 0m/s to 100m/s in 2 seconds, your acceleration would be 50 m/s/s (ie: speed changes 50m/s every second). That would be approx 5 g of acceleration. (50/9.8=5.10 g to be more accurate. Does that help?
@ianthompson4882 So if i state 1 M/S^2 would achieve the speed of 5 meters a second in 5 seconds or 10 meters in 10 seconds or I'm I way off? The time factor is critical, @ 100 m/s in 2 seconds, feels like 5g, but at 8 seconds, it may only feel like 1.3g? So, in the game, these ships are going 1000 m/s. Let's say it takes 15 seconds to get to that speed that's close to 7 Gs whereas two seconds, nets you 51Gs and a really big mess? To further complicate this, let's say we're in space no air, no friction. Nothing to slow us. If I were to fire the thrusters only on the left side of the ship, inducing us to spin right, does this exacerbate the G-force effect? Our initial heading still remains the same, but we're effectively facing in a new direction.
@@kazzdevlin5339 You're on the right track with your 'translational' (straight line) motion examples. ...and you said Maths wasn't your strong suit! ;-) Spinning complicates matters. When you spin, centripetal acceleration is involved. Turning is a form of acceleration, because your velocity is changing (even though its the direction of your velocity, not the 'speed' of your velocity that changes. But ts still a change in velocity, and still involves acceleration) I'm sure you can imagine when you're on one of those amusement rides that spins really fast and you feel pushed outward. That outward 'push' you feel is not actually a force, it's your inertia trying to resist the centripetal (rotational) acceleration. Your centripetal acceleration depends on how your tangential (rotational) speed AND your distance from the centre of rotation. the formula is a=mv^2/r So if you're accelerating in a straight line AND spinning, you need to vector-add the translational (straight line) acceleration to the centripetal (rotational) acceleration. The divide your total acceleration by 9.8 to get the 'g's you'd experience. Does that make sense?
okey okey thats the G's (gravity force) weight - but what im interested in know is how fast in km/h is 1G or 11.6G's worlds fastest Gforce withstand-ed google states that 11.6G's is 35.30394 km/h
Hello, great video. Believe it or not, there are very few of these and this helped a lot. Question, How was the number of 9.8ms^-2 for earth's gravitational acceleration found? Is it just the speed that things accelerate at when they are dropped or is there something else. Also, aren't astronauts technically still experiencing gravity while in space seeing as gravity decreases as the distance from it increases, but it never reaches 0. In order to escape a gravitational pull, don't you have to go an infinite distance away? Great vid, keep up the good work!
Thanks Jamie - so glad it was helpful! 9.8m/s/s (9.8 metres per second per second) is the measured acceleration of an object falling due to the Earth's gravity. It varies slightly depending on your latitude and your elevation. It means that, in freefall, your downwards velocity will increase by 9.8 m/s each second. That is; 9.8 m/s per second = 9.8m^-2. (ignoring air resistance). If you're interested, I have another video that shows my yr10 physics students how to calculate "g" by dropping a lump of plasticine from 1m, 2m, 3m, 4m, 5m and timing it with a stopwatch. Sorry, it's a pretty dry video... lot's of MS-Excel instruction to show how to do the graph. But if you're interested, it's here: th-cam.com/video/K6dMoyM2B-A/w-d-xo.html
Funnily enough I actually already started that one. Haven't finished it yet as I had something else to do, but will try to finish it. Thanks for the quick response btw!
Jamie Parish .... Sorry, I only answered half of your question. So your question about astronauts. Yes! They still experience gravity. In fact even out near Pluto, you'd still experience the Sun's gravity (otherwise Pluto wouldn't orbit the sun!). So some of my scenarios are oversimplifications! The real reason an astronaut is weightless is because when in orbit, he (or she) is actually in freefall 'around' the earth (that's what being in orbit actually is) and objects in freefall experience 0g (as explained in the video).
So, essentially, when you hit the gas pedal in that shiny new hypercar and do 0-100km/h in 2 seconds, you essentially pulled 2.7 horizontal Gs... good shit... although, it's not a constant acceleration, it's more like "descending" acceleration, so, you might be pulling more than 2.7 Gs at first and less than 2 by the time you hit 100 because cars are like that... But, yeah... Sweet.
Hey thanks for your excellent explanation, helped me alot. If I would like to calculate the forces acted on the person in th-cam.com/video/VKcmgPTKLMI/w-d-xo.html Do I need to multiple the g- force by the mass of the person? lets assume he weighs 100Kg, so the person will feel a force of 480.2 N?(0.49*9.8*100=480.2) And about the first example of the rocket, if there was a person in it and there is gravity. How should I calculate the forces acted on the person in this example? becuse the acceleration of the rocket is horyzontal and gravity is vertical.
That was very informative and simple! I wish you were my physics teacher!
Thank you for this absolutely beautiful explanation. This video should be the first one that should come up when looking for 'GForce'! Great video..thank u
I’m watching Formula 1 and G-force is often mentioned and this video helped to understand it. Thank you!
Thank you so much for this video I did not understand g forces at all before watching this.
Really great video! Very easy to understand, nice pace and helpful animation / visuals
Great video, I really enjoyed watching this, thank you for your efforts 😊
Really great explanation!! It was super easy to understand :)) Thank you so much sir
Thank you sir for this video. Extremly helpful! And your animation style is awesome.
astronauts feel weightless because theyre in orbit, basically falling with no end
Thank you very much for simple but great explanations about g-force.
Sir Thank you very much for your simplified explanation. Appreciate your effort.
Had one doubt, feeling lighter as lift comes down (11.03min) is understandable
however while roller coaster comes down (13.24min) why we feel heavier? Is centripetal acceleration different from linear?
ನಾಗರಾಜ್ ರಾವ್ thanks for your comment. On the rollercoaster, it's when you're at the bottom of a dip that you feel heavier. That is, as your vertical velocity slows and you start to go back up again. Sorry if that wasn't so clear!
THANK YOU Ik I’m a year late but this was awesome man
I'm two years late
@@codystuntsgamingandmore4859 I'm 3 years late
Very helpful in understanding *The Expanse* sci-fi TV Series. Have you watched it ?
Wow it was very helpful to understand. Thank for great explanation ❤️
Thanks sir helping me to know what is G-force and you are explaining very nicely 😊
Very helpful and straight explanation
thank you very much this video is very interesting and helped me get the g-force without help
This was help
@@aslydesleek7740 bruh ion even remember what my dumbass wrote😂
Here after Max Verstappen's 51G crash in British GP 2021!
Same here
Thank you for your infomative vdo. Again please confirm whether my understanding is correct.
If I would go above speed of sound 1,100 ft per sec from zero it would mean G force would be 1,100ft/9.8 m act on my body would be around 33 g??
But if I would travel continuously on that speed the inertial would fall to normal. Is my understanding correct?
Pretty close. You are correct that if you travel at a constant velocity, your g-force would be "normal". ie: just 1 g due to gravity.
For your speed of sound scenario, a) you should work in consistent units; preferably SI units. Convert to ft/s to m/s. and b) the g-force on your body would depend on how quickly you accelerated, not your final speed. If you went from 0 to the speed of sound (343m/s) in, say, 10 seconds, that is an acceleration of 34.3 m/s each second, or 34.3 m/s/s, which creates a g-force of 34.3/9.8 = 3.5g. You could then vector add the 1g due to gravity to the 3.5g due to acceleration to determine the overall g-force.
I hope that helps!
Thank you very much, helped me in my thesis.
great < thanks alot
Hi, nice video but i think the fact that you mentioned that astronauts have zero gravity in space is not right. isn't it supposed to be much smaller than that on earth, but not zero?
Hi Irwan, thanks, I'm glad you liked the video. You are correct, even out in space there is gravity, just less and less the further out you go. Interestingly, the reason astronauts are 'weightless' is not because they're away from the Earth's gravity, but because they're in orbit. An object in orbit is actually in freefall, and an object in freefall is weightless. So I probably should have used that as my example. Have a great day!
@@ianthompson4882 thanks alot for he explanation sir
Some sites put G-force=Gs inertial +/- 1, this was for a clothoid loop
Does this mean hypothetically if a lift accelerates downwards at 9,81 m/s2 we would be weightless like in a vacuum? Sorry for my english hope you understand.
Yes, you are correct. That's how astronauts can practice weightlessness in the 'vomit comet' (look it up!).
Btw, you English is just fine! :-)
@@ianthompson4882 I'll definitely search it. Thank you for replying :D
An object at rest must be resisting the rotaional/revolutionary forces of earth or the fluids in the body must be resisting the atmospheric pressure due to gases?
Sorry if this is a dumb question but say you’re only accelerating horizontally, would you not have to + the acceleration due to gravity as it’s acting vertically downwards?
Correct. The total g force would be calculated by the vector addition of the downward and horizontal components, resulting in a 'diagonal" downward and backward g force.
Very good
why is inertial g force in lift is opposite in direction when in roller coaster?
Why is there a inertial g-force upwards in free fall? Great video helped me very well to understand the mathematics
Hi Timo - Any time you have an acceleration, you feel an "inertial force" against the acceleration (again, this is not really a force, but is your inertia resisting acceleration as per Newton's first law. So you can't draw it in a free body diagram. But it *feels* like a force. eg: when you accelerate forwards in a car you 'feel' an inertial 'force' pulling you backwards, even though in a FBD you'd only draw a force pushing you forward).
Now in freefall, you are accelerating downwards, so by my argument above, there should be a felt inertial 'force' upward. But we also feel gravity pulling us downward which cancels out the 'feeling' of the upward inertial 'force'.
I hope that helps.
PS - The other way of analysing g-force is to consider the net force of the surrounds acting on you. That's easier mathematically, (especially for free-fall) but not quite as intuitive. If I get some time, I might create a video showing that approach also.
Thanks for your quick answer Ian, but I still don't understand why there is a force upwards and how is this force called?
@@timo41 There isn't a force upwards (apart from air resistance which is neglected). There is an "apparent force" which is just the body's resistance to the acceleration caused by the downward force of gravity. G-Force is essentialy the apparent weight of the body. The body's inertia results in the the upward G-Force as the inertia is opposite to the downward acceleration.
Damm, i'm so happy that you don't have to explaine here how 2 grown up make chrildern, it will take 5 hour's for you i gess,,
STFU u can't even diss him without a spelling error, the entire point of this was to make is as clear as possible to people who needed it like you
Nadejda Jean Schmidt fuck you idiots
Garold The Great those kind of people has a spider inside their heads
What if the downward acceleration of a falling body is more than "1g"??
If an object is falling, it will naturally accelerate at 9.8m/s/s=1g (or less due if air resistance becomes significant). Never more. However, if an object is pushed downward by some other force with an acceleration greater than 9.8m/s/s, then a negative g force is experienced. For example, if you're in an airplane and it hits turbulence and accelerates quickly downwards, if you don't have your seatbelt on, you may find yourself feeling lifted out of your seat!
Thx dude
So if you travel at ~9.8 m/s upwards, and gravity is pulling you down at ~9.8 m/s, you experience two Gs. But in free fall, you float and feel weightless. Aren't you still being pulled down at 9.8, therefore feel the weight?
Hi Ahmed - When you are in freefall, gravity is pulling you down at 1g AND you are accelerating downward at 1g. So, mathematically, the total G-Force you feel is +1g (due to gravity) and -1g (opposite to your downward acceleration). Which totals zero g.
This matches with what you'd expect if you do a 'thought experiment'. If you were in a lift accelerating downward at 9.8m/s, you would "float" inside the lift, because both you and the lift are accelerating downward at the same rate.
I hope I've understood your question correctly, and my answer helps!
Ian Thompson k thx
Ian Thompson I'm confused about why you measure your body's resistance against the force of acceleration at 9.8m/s squared.
Basically you feel zero gs, acceleration wise. You get pulled down by gravity and also pushed up by inertia. As you float down you feel zero g-force. As he previously mentioned, g-force is the weight you feel because of gravity or acceleration. You will feel zero g-force because of the upward and downward pull of both gravity and inertia. Anything else you feel by free falling is probably because of air resistance which is a completely different topic. You must be familiar with the ball and feather gravity experiment, right?
Hope it clears it up more. 😀
thnxs a lot
God math has never been my strong suit. Ok, the rocker in space is actually relevant to the topic I'm interested in.
I'll explain the issue.
I play Star citizen a space sim game. They've implemented a G force mechanism that causes you to black out. I argue this is unrealistic. Most ship speeds in the game are in meters like 1000 meters.
I'm unaware of whether it's meters per second, meters per second squared, or just meters and hour. (I suspect it's meters per second.)
Seeing as most people black out at 4-5Gs, I figured at 50 m/s squared you're blacking out. However, at 1000 m/s squared, you'd be liquefied. However, this is meters per second squared.
Could you give examples for other units of measuring speed like mph and meters per second?
Hi @kazzdevlin5339. Its not the speed (m/s) that determines the g-force you feel, its the acceleration (m/s/s or metres per second squared). This is, how quickly your speed is changing. If you went from 0m/s (speed) to 100 m/s (also a speed) in, say, 3 seconds, that means your speed is changing at 33.3 metres per second every second. In other words, your acceleration is 33.3 metres per second per second. So that woule be a bit over 3 "g"s of accleration. (33.3/9.8). If you went from 0m/s to 100m/s in 2 seconds, your acceleration would be 50 m/s/s (ie: speed changes 50m/s every second). That would be approx 5 g of acceleration. (50/9.8=5.10 g to be more accurate. Does that help?
@ianthompson4882 So if i state 1 M/S^2 would achieve the speed of 5 meters a second in 5 seconds or 10 meters in 10 seconds or I'm I way off?
The time factor is critical, @ 100 m/s in 2 seconds, feels like 5g, but at 8 seconds, it may only feel like 1.3g?
So, in the game, these ships are going 1000 m/s. Let's say it takes 15 seconds to get to that speed that's close to 7 Gs whereas two seconds, nets you 51Gs and a really big mess?
To further complicate this, let's say we're in space no air, no friction. Nothing to slow us. If I were to fire the thrusters only on the left side of the ship, inducing us to spin right, does this exacerbate the G-force effect? Our initial heading still remains the same, but we're effectively facing in a new direction.
@@kazzdevlin5339 You're on the right track with your 'translational' (straight line) motion examples. ...and you said Maths wasn't your strong suit! ;-)
Spinning complicates matters. When you spin, centripetal acceleration is involved. Turning is a form of acceleration, because your velocity is changing (even though its the direction of your velocity, not the 'speed' of your velocity that changes. But ts still a change in velocity, and still involves acceleration) I'm sure you can imagine when you're on one of those amusement rides that spins really fast and you feel pushed outward. That outward 'push' you feel is not actually a force, it's your inertia trying to resist the centripetal (rotational) acceleration.
Your centripetal acceleration depends on how your tangential (rotational) speed AND your distance from the centre of rotation. the formula is a=mv^2/r
So if you're accelerating in a straight line AND spinning, you need to vector-add the translational (straight line) acceleration to the centripetal (rotational) acceleration. The divide your total acceleration by 9.8 to get the 'g's you'd experience.
Does that make sense?
@ianthompson4882 Moral of the story: If you went from 0 to 1000 m/s doing 7Gs, turning is very bad.
Wanna thank you for all the help much appreciated
okey okey thats the G's (gravity force) weight - but what im interested in know is how fast in km/h is 1G or 11.6G's worlds fastest Gforce withstand-ed google states that 11.6G's is 35.30394 km/h
Hello, great video. Believe it or not, there are very few of these and this helped a lot. Question, How was the number of 9.8ms^-2 for earth's gravitational acceleration found? Is it just the speed that things accelerate at when they are dropped or is there something else. Also, aren't astronauts technically still experiencing gravity while in space seeing as gravity decreases as the distance from it increases, but it never reaches 0. In order to escape a gravitational pull, don't you have to go an infinite distance away? Great vid, keep up the good work!
Thanks Jamie - so glad it was helpful!
9.8m/s/s (9.8 metres per second per second) is the measured acceleration of an object falling due to the Earth's gravity. It varies slightly depending on your latitude and your elevation. It means that, in freefall, your downwards velocity will increase by 9.8 m/s each second. That is; 9.8 m/s per second = 9.8m^-2. (ignoring air resistance).
If you're interested, I have another video that shows my yr10 physics students how to calculate "g" by dropping a lump of plasticine from 1m, 2m, 3m, 4m, 5m and timing it with a stopwatch. Sorry, it's a pretty dry video... lot's of MS-Excel instruction to show how to do the graph. But if you're interested, it's here: th-cam.com/video/K6dMoyM2B-A/w-d-xo.html
Funnily enough I actually already started that one. Haven't finished it yet as I had something else to do, but will try to finish it. Thanks for the quick response btw!
Jamie Parish .... Sorry, I only answered half of your question. So your question about astronauts. Yes! They still experience gravity. In fact even out near Pluto, you'd still experience the Sun's gravity (otherwise Pluto wouldn't orbit the sun!). So some of my scenarios are oversimplifications! The real reason an astronaut is weightless is because when in orbit, he (or she) is actually in freefall 'around' the earth (that's what being in orbit actually is) and objects in freefall experience 0g (as explained in the video).
Intetesting
So, essentially, when you hit the gas pedal in that shiny new hypercar and do 0-100km/h in 2 seconds, you essentially pulled 2.7 horizontal Gs... good shit... although, it's not a constant acceleration, it's more like "descending" acceleration, so, you might be pulling more than 2.7 Gs at first and less than 2 by the time you hit 100 because cars are like that... But, yeah... Sweet.
Oh no jeremy Clarkson
Hey thanks for your excellent explanation, helped me alot. If I would like to calculate the forces acted on the person in th-cam.com/video/VKcmgPTKLMI/w-d-xo.html
Do I need to multiple the g- force by the mass of the person? lets assume he weighs 100Kg, so the person will feel a force of 480.2 N?(0.49*9.8*100=480.2)
And about the first example of the rocket, if there was a person in it and there is gravity. How should I calculate the forces acted on the person in this example? becuse the acceleration of the rocket is horyzontal and gravity is vertical.
Kos
🙏🙏🙏🥰🥰♥️♥️
ooo
I’m in the fifth grade and I don’t know why I’m here