# Flight and the Conveyor Belt **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [February 3, 2006, 12:25am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759 "2006-02-03T00:25:10Z")\ **Posts on this page:** 20\ **Page:** 25
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 5:11pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/481 "2006-02-13T17:11:46Z")
> [@David Simmons](#): > > But I did learn one thing for sure. F certainly does equal ma. As in, “Look, ma, no hands!” ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 13, 2006, 5:13pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/482 "2006-02-13T17:13:59Z")
> [@Colophon](#): > > an irresistible force, such as our putative conveyor belt, would represent infinite energy, and therefore, by Einstein, infinite mass. Actually, no. Under some reasonable thought experiment assumptions (like the existance of friction and wheel mass), the conveyor will provide exactly and only enough force as required to keep the plane stationary. That isn’t equivalent to the irresistable force paradox which, by it’s very nature, requires infinite mass/infinite force. For the plane and treadmill, the plane mass and thrust are clearly finite, so the plane is in no way comparable to an immovable object. The treadmill is then simply scaled to the plane. The force is not infinite. The energy is not infinite. Under some other assumptions (massless wheels, lack of friction), the treadmill _would_ indeed accelerate to infinite speeds. However, it only does that because it _cannot_ deliver sufficient force to keep the plane from moving. This situation might require infinite energy, but it hardly represents an irresistible force. ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 5:37pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/483 "2006-02-13T17:37:35Z")
> [@zut](#): > > Actually, no. > > (snip) Hmm. OK, I take your point. But I do feel that this question is really one of semantics, and the stresses on an aircraft wheel spinning fast enough to hold back the entire aircraft under take-off thrust would be so enormous that they would undoubtedly fail. Plus, would this situation not require zero slippage between the tyres and the conveyor belt? Again, hardly realistic. I maintain that, as far as is possible in the real world, the aircraft would take off. Let’s say that we accept it is possible for the conveyor belt to constantly match the rotational speed of the tyres w.r.t. the ground. The aircraft fires up its engines and releases the brakes. The very instant it begins to roll forward, the conveyor belt starts, thus beginning the positive feedback loop of increasing wheel speed and conveyor belt speed. Eventually, the wheel bearings will overheat and seize up and/or begin to slide over the conveyor belt. Thus the conveyor belt, matching their speed, will slow down and/or stop. Now what? I wonder - is it possible for a plane to reach take-off speed without the wheels turning? Could the engines simply slide the plane along the belt? Maybe. If it could, of course, all the pilot would have to do is lock the brakes, the conveyor belt would come to a complete stop and the plane would slide off down the runway 😛 ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 5:43pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/484 "2006-02-13T17:43:23Z")
Or, basically, what **KeithT** said [here](http://boards.straightdope.com/sdmb/showpost.php?p=7071029&postcount=11) :smack: ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 5:59pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/485 "2006-02-13T17:59:01Z")
Actually, having thought about it some more, I disagree with this part: > [@zut](#): > > Under some reasonable thought experiment assumptions (like the existance of friction and wheel mass), the conveyor will provide exactly and only enough force as required to keep the plane stationary. It cannot do this, because the only way it can provide the force is by _constantly accelerating_. Maintaining a _constant velocity_ – **no matter how fast** – will not prevent the aircraft taking off: the wheels would simply spin faster and the plane would move forwards, as we have seen. So the only way the belt can hold the aircraft stationary is by constantly accelerating. It does not have to accelerate up to a certain point to “balance out” the engine thrust - it has to keep accelerating forever, providing a constant force. This is clearly impossible, and so the plane takes off, even in our theoretical model. Unless of course the belt is massless, in which case how is it imparting force to the wheel? ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 13, 2006, 7:17pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/486 "2006-02-13T19:17:56Z")
> [@Colophon](#): > > Hmm. OK, I take your point. But I do feel that this question is really one of semantics, and the stresses on an aircraft wheel spinning fast enough to hold back the entire aircraft under take-off thrust would be so enormous that they would undoubtedly fail. Again, it all depends on your assumptions. I’ve outlined some, way back in post #106, that match a number of the things you talk about in your post. Nothing wrong with assuming some reasonable, actual material properties and so on, but there’s also nothing wrong with treating this as a thought experiment and supposing the treadmill can accelerate as fast as necessary. > [@Colophon](#): > > It cannot do this [provide exactly and only enough force as required to keep the plane stationary], because the only way it can provide the force is by _constantly accelerating_. > > Maintaining a _constant velocity_ – **no matter how fast** – will not prevent the aircraft taking off: the wheels would simply spin faster and the plane would move forwards, as we have seen. Right. I know that. I’ve been arguing that for the last week. I completely agree. > [@Colophon](#): > > So the only way the belt can hold the aircraft stationary is by constantly accelerating. It does not have to accelerate up to a certain point to “balance out” the engine thrust - it has to keep accelerating forever, providing a constant force. > > This is clearly impossible, and so the plane takes off, even in our theoretical model. In a thought experiment, there’s no reason we have to place an upper bound on belt velocity. We can just let the belt accelerate as much as we want, obtaining whatever velocity we want along the way. Note this _is not_ infinite speed: if the belt accelerates at a constant rate, it never reaches an infinite speed: not today, not tomorrow, not next year, not next century. Clearly impossible for a _real_ system limited by real materials, yes, but not theoritically impossible. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 13, 2006, 7:21pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/487 "2006-02-13T19:21:27Z")
> [@](#): > > Right. I know that. I’ve been arguing that for the last week. I completely agree. Whoops. I agree with your comment that maintaining a constant velocity won’t prevent the plane from taking off (unless you assume some very particular things). I don’t agree that the belt will be unable to provide the necessary force, obviously. ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 7:39pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/488 "2006-02-13T19:39:41Z")
> [@zut](#): > > In a thought experiment, there’s no reason we have to place an upper bound on belt velocity. Yes there is. If the thought experiment is taking place in this universe, then the upper bound is the speed of light. > [@](#): > > Note this _is not_ infinite speed: if the belt accelerates at a constant rate, it never reaches an infinite speed: not today, not tomorrow, not next year, not next century. No, but it will reach _c_ soon enough. > [@](#): > > Clearly impossible for a _real_ system limited by real materials, yes, but not theoritically impossible. Again, I disagree. The “speed of light” barrier is fundamental. It’s not a fuzzy mathematical “infinity” that you can keep accelerating up to and never reach - it’s a concrete value, and if you want to keep accelerating you’re going to have to break it. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 13, 2006, 7:57pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/489 "2006-02-13T19:57:31Z")
Ah, I see. So what happens to inertial forces as the speed of light is approached? Admittedly, this is beyond my field of expertise, but my understanding is that, at near light speeds, the apparent mass becomes larger. In that case, the requred acceleration to maintain a specific force decreases (F=ma, after all), and there’s never any requirement to accelerate past the speed of light–it’s the force that’s important, not the acceleration. So the treadmill can continue to transmit a constant force for an arbitrarily long time. True? Not true? ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 13, 2006, 9:30pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/490 "2006-02-13T21:30:25Z")
> [@1010011010](#): > > Think of it like this: On a line passing thourhg M, oriented 90 degrees to PM, is the point M’. Basically, the force F is pushing the mass M towards M’. PM, PM’, and MM’ form the sides of a right triangle. PM’ is the hypotenuse. I’m not finding any force that would push M back towards P. Only the force pushing it towards M’. PM’ is longer than PM. Before I leave I owe an explanation of why this is wrong (F = ma). In the development of the torque-angular acceleration formula (F = ma) the force F is defined as always being at right angles to the radius line between the mass M and the point P. In order for the line PM to be lengthened there would need to be a component (F = ma) of F directed along PM. The component of F that is along PM is Fcosx, (F = ma) where x is the angle between F and PM. x is **always** 90[sup]o[/sup] by definition and so there is never a component of F directed along PM (F = ma). Your statement assumed that while M moved to M’ the force F (F = ma) retained its original direction. > [@your post above)](#): > > On a line passing thourhg M, oriented 90 degrees to PM, is the point M’. Basically, the force F is pushing the mass M towards M’. But no, F is not pushing M toward M’ (F = ma) on the line which is in the orginal direction of F. What it is doing is maintaining a right angle with PM as it pushes (F = ma). So at M’, at all points between M’ and the original position of M, and at all other points occupied by M the force F (F = ma) is at 90[sup]o[/sup] to PM. Under those conditions there is never any lateral force extending PM (F = ma) and the motion is entirely rotation with no translation at all. But, you say, there can never be such a force. Well, I maintain that a rigid wheel (F = ma) on a rigid conveyor belt which makes line contact with the wheel and applies a force to the rim of the wheel is exactly such a force (F = ma), As I showed, under such a setup the belt (F = ma) can never generate any linear momentum in the wheel (F = ma) and so there can never be a lateral force on the wheel (F = ma). All arguments to the contrary are just addle-headed hand waving. And now one final word. (F = ma) ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 13, 2006, 9:59pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/491 "2006-02-13T21:59:18Z")
> [@David Simmons](#): > > Well, I maintain that a rigid wheel (F = ma) on a rigid conveyor belt which makes line contact with the wheel and applies a force to the rim of the wheel is exactly such a force (F = ma), As I showed, under such a setup the belt (F = ma) can never generate any linear momentum in the wheel (F = ma) and so there can never be a lateral force on the wheel (F = ma). > > All arguments to the contrary are just addle-headed hand waving. > > And now one final word. > > (F = ma) I see. I thought you really did get it, for a while, but I guess I was mistaken. No, **1010011010** is essentially correct: in order for the mass m to accelerate toward P (which is what following a circular path is) there needs to be some force pulling it in that direction. The tension in the string, in this case here. Nothing to do with the tangential force F. This is easy to see by looking at the limiting case: what if the force F is zero? In this case, the mass no longer has an _angular_ acceleration–its tangential velocity doesn’t increase. However, the mass still rotates around the point P, so its total vector velocity _is_ changing, due to the tension in the string. This clearly has nothing to do with the force F, because the force F is zero. Anyway, back to the wheel. The fundamental issue here is that force and acceleration are linked. It’s not really enough to keep repeating it, even in red. You need to _apply_ it. Your contention that an unopposed force does not cause acceleration violates F=ma. That’s hardly addle-headed hand waving. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 13, 2006, 10:23pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/492 "2006-02-13T22:23:29Z")
**David Simmons** , you’re right this is pointless. Find yourself a cylinder and find yourself a treadmill. Place the cylinder on the stationary treadmill, start the treadmill and observe the cylinder flying off of the back. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [February 13, 2006, 10:27pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/493 "2006-02-13T22:27:46Z")
I haven’t had time to slog through this whole other sub-thread/hijack, but will toss in two points: 1. As my department head beat into me when I first started teaching physics, it’s **net** force equals mass times acceleration: [symbol]S[/symbol] **F** = _m_ **a** where both net force and acceleration are vectors. (It’s important to distinguish between “force” and “net force.”) 1. Not sure it this has been discussed already, but have you considered what happens when you apply a tangential force to a disk laying on its side on a frictionless surface? (e.g. a hockey puck resting flat on frictionless ice) (Ans: the disk accelerates linearly and rotationally.) ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [February 14, 2006, 3:22am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/494 "2006-02-14T03:22:23Z")
> [@David Simmons](#): > > Your statement assumed that while M moved to M’ the force F retained its original direction. You seem to be changing your frame of reference when you move from M to M’ to M" etc. and saying “well the force is 90 degrees to PM so its cosine is zero… well the force is 90 degrees to PM’ so its cosine is zero… well the force is 90 degrees to PM” so its cosine is zero." F is 90 degrees to PM. F’ is 90 degrees to PM’. F is _not_ 90 degrees to PM’. The PMM’ triangle is a geometric approximation- kind of like integrating over intervals. As the interval approaches zero the approximation smooths but the overall shape does not change. The spiral will not turn into a circle by using an arbitrarily small interval. > [@David Simmons](#): > > Under those conditions there is never any lateral force extending PM and the motion is entirely rotation with no translation at all. But, you say, there can never be such a force. I say that if you have a point P and a free mass M being accelerated by a constant force F at 90 degrees to line PM, the mass is going to follow a logarithmic spiral, not a circle. Cross reference centripetal force and circular motion. The force diagram is not what you’ve been describing. > [@David Simmons](#): > > All arguments to the contrary are just addle-headed hand waving. The solid disc with the tangential force is probably not analogous to the PMF model we’ve been discussing. If you can show that the two cases _are_ analogous… then what does the spiral motion of the free mass mean in the case of the solid disc (whose radius cannot grow)? ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 14, 2006, 3:48am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/495 "2006-02-14T03:48:00Z")
> [@1010011010](#): > > You seem to be changing your frame of reference when you move from M to M’ to M" etc. and saying “well the force is 90 degrees to PM so its cosine is zero… well the force is 90 degrees to PM’ so its cosine is zero… well the force is 90 degrees to PM” so its cosine is zero." > > F is 90 degrees to PM. > F’ is 90 degrees to PM’. > F is _not_ 90 degrees to PM’. > > The PMM’ triangle is a geometric approximation- kind of like integrating over intervals. As the interval approaches zero the approximation smooths but the overall shape does not change. The spiral will not turn into a circle by using an arbitrarily small interval.I say that if you have a point P and a free mass M being accelerated by a constant force F at 90 degrees to line PM, the mass is going to follow a logarithmic spiral, not a circle. Cross reference centripetal force and circular motion. The force diagram is not what you’ve been describing.The solid disc with the tangential force is probably not analogous to the PMF model we’ve been discussing. If you can show that the two cases _are_ analogous… then what does the spiral motion of the free mass mean in the case of the solid disc (whose radius cannot grow)? 1010011010 (do you mind if I use your decimal alias 1330, it’s shorter?) You are absolutely right and I am wrong and embarassed to boot. :smack: A physical connection to the pivot is required. I want to get my\* mea culpa\* in before someone else comes up with the analysis and claims I copied. :eek: The answer is so simple that I really don’t understand how I screwed it up. I’ll be back. ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [February 14, 2006, 4:10am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/496 "2006-02-14T04:10:24Z")
> [@David Simmons](#): > > 1010011010 (do you mind if I use your decimal alias 1330, it’s shorter?) You are absolutely right and I am wrong and embarassed to boot. A physical connection to the pivot is required. I want to get my _mea culpa_ in before someone else comes up with the analysis and claims I copied. The answer is so simple that I really don’t understand how I screwed it up. I’ll be back. The right-most column is b^0, not b^1… that’s why it’s always the “ones” column regardless of the base: b^0=1. Still, 1010011010 **0** = 1332… not 1330. So do you think that the PMF scenario is not representative of the disc-with-tangential-force scenario, or are you reconsidering the linear acceleration of an ideal disc on an ideal conveyor? ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 14, 2006, 6:12am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/497 "2006-02-14T06:12:46Z")
> [@1010011010](#): > > The right-most column is b^0, not b^1… that’s why it’s always the “ones” column regardless of the base: b^0=1. Still, 1010011010 **0** = 1332… not 1330. > > So do you think that the PMF scenario is not representative of the disc-with-tangential-force scenario, or are you reconsidering the linear acceleration of an ideal disc on an ideal conveyor? I’m starting from scratch. The conveyor exerts a force on the tire in the direction of the belt travel and this force is applied to the hub moving the plane backward. The most force it can exert is the sum of the tire resistance, the hub friction and the acceleration of the wheel. In a normal takeoff the plane also has to overcome all of those. However with the belt going backward at the same speed as the plane is going forward relative to the ground the hub friction and the wheel acceleration are both increased. How much I don’t know but I think I can make some assumptions and even if they are a little off a comparison between the belt takeoff and a normal takeoff can be approximated. I think the tire resistance is probably pretty constant with velocity but I’m willing to accept that it is some function of velocity if anyone knows what that is. It’s not really a difficult problem if you have the data, but it gets a little sticky if your just making what you hope are reasonable guesses. I don’t want to question your knowing your own post name, but why is it 1010011010 at the start of your post and 1010011010 **0** later on? ---
**Author:** ![MonkeyMensch](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)\ **Post date:** [February 14, 2006, 7:50am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/498 "2006-02-14T07:50:19Z")
> [@David Simmons](#): > > I’m starting from scratch. The conveyor exerts a force on the tire in the direction of the belt travel and this force is applied to the hub moving the plane backward. The most force it can exert is the sum of the tire resistance, the hub friction and the acceleration of the wheel. In a normal takeoff the plane also has to overcome all of those. However with the belt going backward at the same speed as the plane is going forward relative to the ground the hub friction and the wheel acceleration are both increased. How much I don’t know but I think I can make some assumptions and even if they are a little off a comparison between the belt takeoff and a normal takeoff can be approximated. I think the tire resistance is probably pretty constant with velocity but I’m willing to accept that it is some function of velocity if anyone knows what that is. Well I guess now would be an appropriate time to follow up on my previous post. It turn out that rolling resistance increases with speed. My first guess was that resistance increased linearly with speed. It turns out, however, that it increases with the square of speed. From [this site](http://www.ctts.nrel.gov/analysis/pdfs/itec02_modeling_paper_final.pdf) > [@](#): > > In April of 2002, NREL released ADVISOR 2002. In this > latest release, an updated rolling resistance model was > included along with related tire rolling resistance test > data based on the SAE J2452 test procedures. Previous > to the ADVISOR 2002 release, the road-load force due > to rolling resistance was calculated as: > Fr = Cr × m × g > Where: Cr is the rolling resistance coefficient > m is the vehicle mass > g is the gravitational constant. This is the relationship that some people, I believe, were thinking of when stipulating that rolling resistance is a constant with respect to speed. However, > [@](#): > > Note that this simplified rolling resistance force is not > dependent on vehicle speed and does not account for > changes in tire pressure. > The improved rolling resistance model included in > ADVISOR 2002 is based on the following equation: > Fr = P[sup]α[/sup] L[sup]β[/sup] (a + bV + cV [sup]2[/sup] ) > Where: > P is the tire pressure in MPa > L is the tire load in kg > V is the vehicle speed in m/s > α, β, a, b, and c are coefficients used to > fit the experimental rolling resistance > data. I’ve seen graphs for rolling resistance at low speed (\<30 moh) and the curve _is_ slightly dished, concave upwards. It’s not a huge curve, however, and explains why a constant resistance is a decent enough approxamation for low speeds. But that squared term in the above equation means that at high speed the resistance is going to become, well, biggish. So it turns out that a steady state arrangement of the airplane being held back solely by frictional forces on a non-accelerating treadmill is predicted by classical physics. Of course, if you think that blown tires violate the spirit of the experiment I would only add that if the tires blow, the airplane **never** takes off. 😉 ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 14, 2006, 12:30pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/499 "2006-02-14T12:30:24Z")
> [@MonkeyMensch](#): > > I’ve seen graphs for rolling resistance at low speed (\<30 moh) and the curve _is_ slightly dished, concave upwards. It’s not a huge curve, however, and explains why a constant resistance is a decent enough approxamation for low speeds. But that squared term in the above equation means that at high speed the resistance is going to become, well, biggish. > > So it turns out that a steady state arrangement of the airplane being held back solely by frictional forces on a non-accelerating treadmill is predicted by classical physics. Of course, if you think that blown tires violate the spirit of the experiment I would only add that if the tires blow, the airplane **never** takes off. 😉 However, note that: “α, β, a, b, and c are coefficients used to fit the experimental rolling resistance data.” I’m fairly certain that the _form_ of the equation is something of a best-guess, and in any case it’s based on matching experimental data as opposed to fundamental physics and material properties. There’s nothing particularly _wrong_ with matching experimental data, except that in this particular… um… application, we’d be extrapolating the relationship up to velocities that are orders of magnitude beyond where the original data are. Not unreasonable if that’s the best approximation available, but somewhat tenuous. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 14, 2006, 12:43pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/500 "2006-02-14T12:43:58Z")
(Although, after a bit of thought, I believe **David** is working the problem as it was originally stated in Cecil’s column: the treadmill matches the plane fuselage speed, so the wheels spin essentially twice as fast as they would normally. That’s not such a stretch as the continuously-accelerating case–particularly since the plane wheels are larger diameter that a car’s–so the experimentally-fit line inspires more confidence.) [Previous page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=24) [Next page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=26)