Acceleration of aircraft carrier take-off | One-dimensional motion | Physics | Khan Academy
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Acceleration of aircraft carrier take-off | One-dimensional motion | Physics | Khan Academy

September 8, 2019

So I’m curious about how much acceleration does a pilot, or the pilot and the plane, experience when they need to take off from an aircraft carrier? So I looked up a few statistics on the Internet, this right here is a picture of an F/A-18 Hornet right over here. It has a take-off speed of 260 kilometers per hour. If we want that to be a velocity, 260 km/hour in this direction, if it’s taking off from this Nimitz class carrier right over here. And I also looked it up, and I found the runway length, or I should say the catapult length, because these planes don’t take off just with their own power. They have their own thrusters going, but they also are catapulted off, so they can be really accelerated quickly off of the flight deck of this carrier. And the runway length of a Nimitz class carrier is about 80 meters. So this is where they take off from. This right over here is where they take off from. And then they come in and they land over here. But I’m curious about the take-off. So to do this, let’s figure out, well let’s just figure out the acceleration, and from that we can also figure out how long it takes them to be catapulted off the flight deck. So, let me get the numbers in one place, so the take-off velocity, I could say, is 260 km/hour, so let me write this down. So that has to be your final velocity when you’re getting off, of the plane, if you want to be flying. So your initial velocity is going to be 0, and once again I’m going to use the convention that the direction of the vector is implicit. Positive means going in the direction of take-off, negative would mean going the other way. My initial velocity is 0, I’ll denote it as a vector right here. My final velocity over here has to be 260 km/hour. And I want to convert everything to meters and seconds, just so that I can get my, at least for meters, so that I can use my runway length in meters. So let’s just do it in meters per second, I have a feeling it’ll be a little bit easier to understand when we talk about acceleration in those units as well. So if we want to convert this into seconds, we have, we’ll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I’ll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you’re doing any type of this dimensional analysis, you really should see whether it makes sense. If I’m going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that’s why we’re dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that’s why we’re multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 divided by 3600 meters per second. So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I’ll just round it to 72, because that’s about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let’s think about what that acceleration could be, given that we know the length of the runway, and we’re going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let’s think a little bit about it. The total displacement, I’ll do that in purple, the total displacement is going to be equal to our average velocity while we’re accelerating, times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It’s going to be our final velocity, plus our initial velocity, over 2. It’s just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we’re trying to get to 10 m/s, or we’re trying to get 10 m/s faster, and we’re accelerating at 2 m/s squared, it’ll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to change in velocity over change in time. You multiply both sides by change in time, and you divide both sides by acceleration, so let’s do that, multiply both sides by change in time and divide by acceleration. Multiply by change in time and divide by acceleration. And you get, that cancels out, and then you have that cancels out, and you have change in time is equal to change in velocity divided by acceleration. Change in velocity divided by acceleration. So what’s the change in velocity? Change in velocity, so this is going to be change in velocity divided by acceleration. Change in velocity is the same thing as your final velocity minus your initial velocity, all of that divided by acceleration. So this delta t part we can re-write as our final velocity minus our initial velocity, over acceleration. And just doing this simple little derivation here actually gives us a pretty cool result! If we just work through this math, and I’ll try to write a little bigger, I see my writing is getting smaller, our displacement can be expressed as the product of these two things. And what’s cool about this, well let me just write it this way: so this is our final velocity plus our initial velocity, times our final velocity minus our initial velocity, all of that over 2 times our acceleration. Our assumed constant acceleration. And you probably remember from algebra class this takes the form: a plus b times a minus b. And so this equal to — and you can multiply it out and you can review in our algebra playlist how to multiply out two binomials like this, but this numerator right over here, I’ll write it in blue, is going to be equal to our final velocity squared minus our initial velocity squared. This is a difference of squares, you can factor it out into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what’s really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, sometimes called distance, if you’re just using the scalar version, and really we are thinking only in the scalar, we’re thinking about the magnitudes of all of these things for the sake of this video. We’re only dealing in one dimension. But sometimes you’ll see it written like this, sometimes you’ll multiply both sides times the 2 a, and you’ll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it’ll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it’s not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you’re just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we’ve just derived ourselves a kind of a neat formula that is often not derived in physics class, but let’s use it to actually figure out the acceleration that a pilot experiences when they’re taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that’s 80 meters, times 80 meters, is going to be equal to our final velocity squared. What’s our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it’s just going to be minus 0 squared, which is just going to be 0, so we don’t even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let’s just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared over 2 times 80 meters. And what we’re gonna get is, I’ll just write this in one color, it’s going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we’re squaring the units, and then we’re going to be dividing by meters. So times, I’ll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we’re going to get is this meters squared divided by meters, that’s going to cancel out, we’re going to get meters per second squared. Which is cool because that’s what acceleration should be in. And so let’s just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don’t want to forget about this part right over here. 72 squared divided by 160. So we have, and we can just use the original number right over here that we calculated, so let’s just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven’t done a lot on force yet, we’ll talk about this in more depth, is that this pilot would be experiencing more than 3 times the force of gravity, more than 3 g’s. 3 g’s would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he’s, you know, if this is the chair right here, his pilot’s chair, that he’s in, so this is the chair, and he’s sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let’s say this is you, this is your feet, and this is your face, this is your hands, let me draw your hands right here, and if you had essentially two more people stacked above you, roughly, I’m just giving you the general sense of it, that’s how it would feel, a little bit more than two people, that squeezing sensation. So his entire body is going to feel 3 times heavier than it would if he was just laying down on the beach or something like that. So it’s very very very interesting, I guess, idea, at least to me. Now the other question that we can ask ourselves is how long will it take to get catapulted off of this carrier? And if he’s accelerating at 33 meters per second squared, how long would it take him to get from 0 to 72 meters per second? So after 1 second, he’ll be going 33 meters per second, after 2 seconds, he’ll be going 66 meters per second, so it’s going to take, and so it’s a little bit more than 2 seconds. So it’s going to take him a little bit more than 2 seconds. And we can calculate it exactly if you take 72 meters per second, and you divide it by 33, it’ll take him 2.18 seconds, roughly, to be catapulted off of that carrier.

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  1. great job showing the concept instead of just telling the youngsters formula's! love the physics vids. please do computer science soon khan… i would love to see some oop languages!

  2. Khan, pls pls relativity theory i cannot wait to see u explain it! and if ur time permits, pls explain some of the quantum mechanics concepts. dude, I love ur soul ^^ Salam

  3. This doesn't take into account the fact that jets on aircraft carriers take off into the wind, and the carriers themselves are also traveling forward with respect to the water.

  4. m8tate is correct, the jet needs an >airspeed< of 260km, so you should have subtacted the wind speed and ship speed from 260. Sill, its a fun problem.

  5. Thats not the runway. the runway is that slightly slanted one just to the right of that communications tower.

  6. I think the actual acceleration is a little higher by the end of the runway (which as mignik01 said is the ramp by the tower) due to the fact that I think the pilot often reaches take-off velocity before reaching the end of the ramp.
    I don't think the wind comment was very fair though (m8tate). Educational videos on YouTube can't be expected to successfully tutor people on mechanics whilst also involving friction, unless assumed to be constant.

  7. @JimNtexas Actually, @m8tate is wrong, the ship speed won't influence the speed needed for takeoff. And about the wing resistance, we can't calculate that unless you got the aerodinamical characteristics of a F/A 18 Hornet.

  8. @DekraDaFurry – While the airspeed needed to fly is what it is, in order the takeoff roll is definitely influenced by the wind speed over the deck (V0 in Sal's equation).

    It's highly unlikely that the F-18 could make a successful takeoff if the ship was stationary with no wind. The cat probably couldn't accelerate the jet to takeoff speed and it would fall into the drink.

    Try this thought experiment. Could the F-18 make a successful takeoff if the ship was going 100 knots in reverse?

  9. Don't you also have to consider the ships speed? It has an initial velocity of approx. 50kmh which is significant making the planes initial velocity of 50kmh if the ship is traveling forward in parallel with the runway.

  10. @JimNtexas Yes it could but only if there were a 200 knot wind traveling in the same direction as the ships travel(backward.).

  11. you sersly do not know any thing about these aircraft. you dont have to be me to know that you cant take off if theres planes in the way that strip with no plane is were they land and take off [email protected]#$

  12. @TheNalu7 shut up dumbfuck I hope you crash your fucking ass with your plane skills in afganistan fighting a pointless war. That is all…

  13. v sub i should be at least 15 m/s – this is after all taking off from an aircraft carrier which would be moving at a good speed not to mention turned into the wind.

  14. I was like no way, so it takes off in about 1 second so I YouTubed it and it did take off in about 1.5 seconds.

  15. We needed it bec-… Wait, 1 month old? What the heck. All we knew was how far the plane went, and how fast it needed to be going by the end of it. If the distance was 160 meters (twice as far) then the plane could accelerate at one half the speed (16.5 m/s^2).

  16. I recommend to use this formula x-x0 = 1/2(v0+v)t and then substitute 80m for x-x0 and 72m/s for v to get t = 2.22 s.Substitute the value of 't' and 'v' in this formula which is v = v0 + at to get a approx = 33m/s^2.It's that simple.

  17. Wait wait, I'm only 1:32 into the video, how are you able to use your mouse pointer like that in writing? You must have an amazing manual dexterity. ^5

  18. they take off and land on the same runway track. The place you said where they take off is plane holding area. just thought I would let you know 🙂

  19. Just want to point out none of these calculations account for drag, which is an extremely strong opposing force.

  20. Its not accurate as many already said, to takeoff and landing the carrier must be moving to reduce the airplane velocity…
    So, V0 is arround 50 km/h or 30 knots

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