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Homework Help: Simple Aerodynamics Question

  1. Sep 27, 2006 #1
    Hey, first post. I've been searching around everywhere but I am yet to come up with an an equation or theorem that relateds altitude to speed and the forces that act on a plane (thrust, weight etc). I am perhaps assuming that, in at least this question, the altitude of the plane doesn't affect it's speed if the forces stay the same? Anyway here it is

    A business jet is traveling at a constant speed of 150 m/s while its engines provide a total thrust of 25 kN

    a) If it is in level flight, what is the magnitude of the total drag on the jet? (I answered with 25 kN because since it was travelling at a constant speed and in level flight, drag = thrust)

    b) If the jet reduces its altitude by 500m, what would be the new maximum possible speed of the jet? (This is the one I'm stuck on. The question doesn't specify the actual altitude of the plane so I've come to think that it is a bit of a trick question and the max speed is still 150 m/s because the altitude is irrelevant if the forces stay the same?)

    Hopefully someone can help me out with question b)...and question a) if I buggered that one up too.
  2. jcsd
  3. Sep 27, 2006 #2


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    Staff: Mentor

    Welcome to PF, Dave. I'm no aero expert (by any means), but I'd start by calculating the difference in air density for each 500m change in altitude. If it's a constant multiplier (I don't know if it is, but it's a possibility), then use that multiplier to slow the max speed by the same ratio. Air resistance has several non-linear terms, I believe (again, I don't know the details), but presumably at least the first linear term would be a simplified answer to this question. Do you have a way to check this answer?
  4. Sep 27, 2006 #3

    a is correct because if there was either more or less drag than thrust the plane would be accelerating (slowing or speeding up).

    b however is not correct. True the forces stay the same, but the air is thicker at lower altitude so the plane will move slower since with denser air the same amount of drag can be generated at lower speeds. The author or your instructor is probably assuming a constant change in air density. Look for a formula relating air density with altitude and air density with drag.
  5. Sep 27, 2006 #4
    Thanks for the help guys. Okay I've done what you said berkeman by calculating the air density and lift value for 500 m increments using a handy tool on the nasa website http://www.grc.nasa.gov/WWW/K-12/airplane/density.html

    I plotted the values into Excel and the graphs of altitude vs density AND altitude vs lift were both linear. I'm not sure how to deal with it now though. I somehow need to get some values for drag here which in this case equal the thrust values and then i need to work out the new max speed.

    On the nasa site (link above) it said that if you halve the density you halve the drag. The problem here is I'm not given a starting altitude, I'm simply told to reduce it by 500m. The equation given for density vs drag on the nasa site is Drag = Constant * Density.

    Hopefully someone can point me in the right direction here.
  6. Sep 27, 2006 #5
    Oh and here is the attached excel spreadsheet with the graphs I made about intervals of 500 m as suggested by berkeman. Hopefully it is of some use.

    Attached Files:

  7. Sep 28, 2006 #6

    Anyone got any help for this? I still haven't been able to find an answer yet I don't think it'd actually be that hard. I was assuming a simple formula would do the trick as the rest of the questions on the sheet were relatively easy.
  8. Sep 29, 2006 #7


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    Here's what I would do as a first shot since this question statement seems to be missing a lot of info:

    Calculate the percentage change in density after decreasing 500m. You have this as the slope of the density-altitude plot (it helps a lot that it is linear). Then simply multiply the top speed by the inverse of that percentage. The density should increase, say by 10%. That would mean multiply the top speed by (110%)^-1.

    It's silly, but since the actual problem is much more complicated than that, this seems like what they may want you to do.
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