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Homework Help: Acceleration of an aircraft

  1. Jul 29, 2009 #1
    Hi, I don't know how hard this problem really is, I've tried setting up a differential equation and get garbage answers. Can anyone help me?

    Imagine two identical aircraft with identical drag, wing area and thrust but different payloads, such that one has a max straight and level speed of 205mph and the other 210mph. Initially, at t=0 the slower aircraft is 200ft vertically above the faster aircraft, and then starts to dive at a CONSTANT RATE in the most efficient way (I'll return to what I mean by this in a minute) such that at some point ahead of the starting point, it ends up at the same height as the faster aircraft.

    Clearly, immediately after t=0, the faster aircraft will be ahead of the slower one when observed from above or below, but as the slower aircraft descends, it will accelerate such that it MAY overtake the faster one. Its speed willl increase until it tops out at some new higher speed. Once it levels out, it will start to slow again back towards its initial speed of 205 mph.

    By "most efficient way", I mean that the slower aircraft uses its excess potential energy such that it stays ahead of the faster one for the longest time possible after t.

    The questions I'm trying to answer are: what is the max speed the slower aircraft will reach? How far will it have travelled when it reaches this speed? Will it be ahead of or behind the faster aircraft at this point and by how much ? For a given distance x from the starting point that is beyond the point at which the slower aircraft levels out, how do I plot the relative positions of the two aircraft?

    This is actually a real world problem that came about in an air race. If you're interested, I'd be happy to tell you about it, its quite fun!


  2. jcsd
  3. Jul 29, 2009 #2
    This problem makes me think of the brachistochrone problem from variational calculus.
    I guess a modified version of it could help with plotting the position of the plane.

    Last edited by a moderator: Aug 6, 2009
  4. Jul 29, 2009 #3


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    Welcome to the PF. Interesting problem.

    I'm not sure that the top aircraft will overtake the lower aircraft, however. It may gain speed in the dive, but does its horizontal speed actually increase, or just its diagonal diving speed?
  5. Jul 29, 2009 #4
    I think we can just consider this as a typical brachistochrone problem in a frame moving at 205mph? I think I'll give this a go when I get home.
  6. Jul 30, 2009 #5
    Thanks for the replies guys, I didn't expect such enthusiastic responses so quickly!

    Berkeman - With regard to the question of whether the top plane overtakes the slower one, this is a real world situation that occurred in a handicapped aircraft race last week. The question I'm really trying to answer is whether one of the aircraft was cheating. From a common sense point of view, it seems unlikely that a 5mph slower aircraft could gain enough speed to overtake a faster one, given that the only real advantage is a small extra amount of potential energy from the excess height. Given that both aircraft are initially travelling at their max straight and level speed, I would have thought that drag would have eaten up most of the excess energy as the plane descended.

    In the general case, it may be that the slower aircraft doesn't ever overtake the faster one, but clearly by choosing, lift and cross sectional areas appropriately, you could create a situation where this is acheiveable.

    If anyone can help with this, it will be very much appreciated, as it will enable us to build it into our handicapping process which will benefit the sport of handicap air racing.

    QueenOfBabes- appreciate your time looking at this!

    Be gentle on me with the maths though - I was an engineering post grad 20 years ago but its a little rusty!


  7. Jul 30, 2009 #6
    Guys, I've just looked up the Brachistochrone problem, and I think this problem is a bit different. The Brachistochrone problem doesn't consider aerodynamic drag, which is going to be the dominant speed limiting factor here.

    Remember, we are talking about aircraft which are initially at an equilibrium constant maximum straight and level speed because their thrust from the engine exactly matches the drag on the aircraft, which is proportional to v^2. You then add energy to the forward motion by decreasing potential energy due to its height, so the drag increases. Without drag, the speed increase is huge, which clearly isn't what happens in the real world.

    Hope that gives a clearer definition of the problem. If anyone needs it, the drag is given by

    Fd = (Cd.r.A.v^2)/2

    where Cd is the co-efficient of drag which is known, r is the known mass density of air, A is the frontal area of the aircraft (known) and v is the velocity.

    thanks again!
  8. Jul 30, 2009 #7


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    Well if you want to include drag. As the aircraft descends its cross sectional area in the horizontal direction increases by some amount.

    Now the way I would do this analysis is by using the simple case of neglecting drag. Then find time 't' that it would take to descend 200ft as a function of descent angle. Then use that time and compare horizontal distances both planes have traveled in that time.

    After this one can add the drag term.

    However, we don't know by how much the area in the horizontal direction of the descending craft increases. But drag has a V^2 term and V is large in both cases (205)^2 and (210)^2. Also since these are sport craft their fuselages and tails should be quite slender, so I would assume any change in area has little effect on the drag force so I would assume we can say drag force remains relatively constant for both craft.

    My first impression is that the plane descending will not catch the faster one.

    Edit: On second thought the horizontal drag force will have some component of 205mph i.e. [tex](205 cos\theta)^2[/tex] where theta is the descent angle. I still doubt this is enough to overtake the faster plane
    Last edited: Jul 30, 2009
  9. Jul 30, 2009 #8


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    Oh and I also assumed the plane kept the same velocity during its descent and has some different horizontal V.
  10. Jul 30, 2009 #9
    You're right: including the drag term into the standard brachistochrone analysis left me quite stuck.
  11. Jul 30, 2009 #10
    With both aircraft equal other than specified, drag should be set proportional to velocity squared.

    The proportionality constant, itself doesn't have to be known; only that D1 / D2 = V12 / V22.

    This disregards complicatating factors such as reduction in drag coefficient as the turbulent flow regime lasts longer down the cord of the wing, or along the body of the aircraft with velocity increase, as I don't think any have a significant effect.

    The efficency of the engine/propeller vs. velocity seems equally disregardable; you think? At best we might be talking about a 10% increase in velocity, max maybe.
    Last edited: Jul 30, 2009
  12. Jul 31, 2009 #11
    QueenOfBabes - you're not the only one!

    djeitnstine - I don't think you need to consider an increase in cross sectional area in the vertical plane. Remember the plane will be moving such that the cross sectional area relative to the airflow will remain constant - the velocity vector will just have changed direction to be angled down. The velocity of the aircraft will not be a constant as it dives, it will accelerate until it reaches a new equilibrium state. At Phrak point out, the drag term is proportional to V^2, so what we're really trying to do is say is:

    "okay, if the plane descends, it trades Potential Energy over some time T for increased velocity, such that the energy per unit time is exactly balanced by the increase in drag (given by the equation stated)."

    Anyone know how to do this?

    Phrak - it sounds like you're on the right lines and you know a lot more about fluid dynamics than I do. Keep upi the good work!
  13. Jul 31, 2009 #12
    I wouldn't try to solve the various problems all at once.

    Solve for the airplane flying straight and level up to time t0, then immediately the momentum is directed at a downward angle, theta for a change in Y feet of altitude.

    The problem statement might be "what is the gain in distance X after diving Y feet at angle theta should the plane have remained flying straight and level?"

    But I left out the part where the airplane is still traveling faster after havingleveled-off at the bottom of the curve.

    Or am I way behind on this and this has already been identified as the central problem?
  14. Aug 3, 2009 #13
    I think that's probably right, Phrak. I just don't know how to do it.

    I think you can ignore the levelling out part, assume that the plane continues to dive until the point where it either overtakes the faster aircraft, or runs out of its height advantage. those are your boundary conditions.
  15. Aug 3, 2009 #14

    I understand your problem, but the slower airplane simply isn't going to have enough energy to catch the faster one in any sensible scenario.

    If both airplanes are at maximum speed, then thrust is just balancing drag. Drag builds proportionally to V squared (as someone has stated above), which at the speeds you suggest are considerable. 150ft isn't enough energy to add any sensible amount of kinetic energy when the additional drag has been paid for.

    The most efficient way to use the excess energy is as slowly as possible, because then the increased drag is minimised. In this case, you're not going to be able to increase speed anything like 5mph and the faster aircraft keeps accelerating away.

    I can't write down an analytic solution for you, but I can tell you I have run the problem on our computational fluid dynamics model for Cd values from 0.2 to 0.6, which bounds all light aircraft I'm aware of, and unless the aircraft is above 30,000ft (to reduce air friction) or you add available energy (either by a step lightening of weight or by increasing thrust), an overtaken airplane won't catch the overtaker again, period.

    It'll be interesting to see someone write down the equation for the relative motion though. Its sure an interesting problem to try to solve explicitly. I'll have a go if I get time.


  16. Aug 3, 2009 #15
    To be clear, the Cd values I've used are referenced to frontal area not wing chord, ie on the basis that automotive engineers use, not the airspace system. The reason I did this is a) we're more interested in frontal area than lift performance and b) most of my research is on automobile drag, and I'm just more used to the equations in this form.
  17. Aug 3, 2009 #16
    I dunno. This makes for a more realistic analysis--but is harder for a first attempt. The diving plane A, if initially behind will exceed the speed of the plane B, flying straight and level if it is to win. So at the bottom of it's dive it still has excess speed.

    The boundry condition to establish is a tie; each cross the finish line at the same moment. I can't tell as yet, if the velocities have to match. I don't think so. Given an initial, excess hight of Y, both the the optimum angle of decent and when to begin the decent are unknows.
  18. Aug 5, 2009 #17
    I'm pretty certain the speeds DON'T have to match, Phrak. Imagine a scenario where the 2nd plane had no drag (I agree, very unrealistic!!). then all of its PE could be converted to KE, and the velocity gain would be huge v=sqrt(2gh2), or sqrt(2x9.8x50) if we work in SI units, or 30 or so metres per second, or 60mph.

    IF a speed increase of this size was possible, then the slower plane could be a long way behind and rapidly accelerate past the faster one as they approached the line.

    Except of course most of the excess PE is going to get used overcoming increased drag rather than increasing V!!
  19. Aug 8, 2009 #18
    I think you guys are making this problem harder than it needs to be! First off, determine the maximum speed the slower aircraft can accelerate to using the height advantage. If that's less that 210mph, then you don't need to worry about solving the PDEs.

    Heres my thinking:


    Thrust = Drag since the aircraft isn't accelerating.

    Thrust = rho.V1^2.Cd.A where v1 =205mph (1)

    When the aircraft starts to descend, the horizontal component of the resultant force F=m.g.cos(theta) where theta is the angle between the horizontal and the downward path of the aircraft. We don't know this angle, and can try some different values out. As you will see, the actual result is pretty insensitive to it. If we assume the airplane descends 200ft over 1 mile, then theta = arctan(200/5280). This is a small angle, and if the distance travelled is bigger, then the angle will be even smaller. Fortunately, cos(theta) doesn't change much for small angles so the force won't vary much, but we can at least calculate some limiting values.

    The aircraft will accelerate on its downward path until it reaches some new equilibrium value of V, say V2, at which point the drag will equal the thrust of the aircraft plus the increased force m.g.cos(theta). At the end of the dive, the aircraft will slow from V2 back to V1.


    Thrust +m.g.cost(theta) = rho.V2^2.Cd.A in the new, faster equilibrium (2)

    Substituting for thrust from (1) into (2) and rearranging

    m.g.cos(theta) = rho.(v2^2-v1^2).Cd.A

    From which

    V2^2 = m.g.cos(theta)/(rho.Cd.A) - v1^2. (3)

    Now, you don't tell us what kind of aircraft these are, but if they do 200mph they are pretty fast. I've assumed Glassairs, because I don't know much that is faster or less draggy. If a Glassair can't catch faster Glassair, then probably nothing can.

    Subsituting in the numbers (see the Glassair datasheet for values of Cd and A - and remember to convert to the vertical plane of reference, aircraft Cd values are given referenced to wing area), v2= 208.69mph.

    So since the slower aircraft can't catch the faster one, we don't need to bother solving the full set of equations of drag motion. The slower one can't catch the faster one - unless its a lot less draggy than a Glassair - and I mean A LOT! The Cd would need to be less than half the Glassair value.

    Do I win a cookie?
  20. Aug 8, 2009 #19


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    The horizontal component is actually F=mgsin(theta), which means that the final answer is very dependent on the value of theta.
  21. Aug 8, 2009 #20
    Fluid, Thanks so much for posting that - very useful!

    Ideasrule - I think Fluid is right, it is cos of the angle. The horizontal distance Fluid gives is the adjacent, the vertical distance is the opposite. theta is arctan opp/adj. The increase in speed along the hyp is given by cos (theta).


  22. Aug 8, 2009 #21


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    The equation Thrust +m.g.cost(theta) = rho.V2^2.Cd.A would imply that the smaller theta is, the larger the mgcos(theta) term has, which can't be right because the term equals zero when theta=0.
  23. Aug 9, 2009 #22
    I think you guys are arguing over which way you define the frame of reference. Draw it out and you'll see you're talking about calling different angles theta!
  24. Aug 10, 2009 #23
    craigbeevers - if you're defining the angle as you do, then ideasrule is right. I looked at your post again and see you're defining it the same way as ideasrule. If you used cos, then you'd get a speed of over 300mph, which is clearly garbage!
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