I have a question about the characteristic dimension.

Q1 The drag on an airship is known to depend on the air density, ρ , dynamic viscosity, μ , speed relative to the air , V , and a characteristic dimension, L . An airship is to operate at 20 m/s in air at standard conditions. A model is constructed at 1/20 scale and tested in a wind tunnel at the same air temperature to determine drag. (a) Perform a dimensional analysis to determine the criterion which should be considered to obtain dynamic similarity.

like the question said for the characteristic dimension L, How can i know what dimension is L.

  • $\begingroup$ L will be a length... will it be the diameter or the length from nose to tail? $\endgroup$
    – Solar Mike
    Nov 22, 2018 at 21:18
  • $\begingroup$ can you explain why is it a length, because i do not really understand what a characteristic dimension even after going through some online note. $\endgroup$
    – josh
    Nov 22, 2018 at 21:20
  • $\begingroup$ Length as in distance between two points. $\endgroup$
    – Solar Mike
    Nov 22, 2018 at 21:21
  • $\begingroup$ In your case, a characteristic length is any length that allows you to reconstruct the airship to scale. Lets say you know the length of the pilot's nose full scale. Given a scaled length, you can build the entire model to scale by keeping the same proportions. In this kind of scenario, the length is truly arbitrary. $\endgroup$
    – Phil Sweet
    Nov 26, 2018 at 1:34
  • $\begingroup$ Depending on the age of the pilot, their nose might actually grow during the course of a flight... $\endgroup$ Nov 27, 2018 at 23:57

1 Answer 1


The drag will actually depend on the frontal area. This can be calculated as $constant*L^2$. What the constant is depends on what you select L to be.

The question is concerned specifically with the dimensionality, i.e. how does the drag force scale as you vary each of the components in the equation?

In this example, you're looking at a 1/20th scale model. A full scale model will be 20 times as long, 20 times as wide, have wheels 20 times as big, and consequently, the frontal area will be 400 times larger.

It doesnt matter what you select as “L” since the constant is determined experimentally. I’d suggest something easy to measure, like overall length of the ship.

If, in this particular model, the length is 4 times the diameter, and you define L=diameter, then the constant that is measured experimentally would simply be 1/4 of the value it would have been if you defined L=length.

  • $\begingroup$ Does the overall length of the ship always control the frontal area? That would be the diameter? $\endgroup$
    – Solar Mike
    Nov 23, 2018 at 6:36
  • $\begingroup$ If the full scale ship is the same shape as the model, then any length will do. If the ship has a radius of 2, and a length of 4 so a frontal area of pi*r^2 = 4pi, could also be calculated as (pi/4)L^2 = 4pi Only the constant has changed from pi to pi/4. If the lengths double, pi*4^2 = 16pi and (pi/4)*(8^2)=16pi $\endgroup$ Nov 23, 2018 at 9:24
  • $\begingroup$ ^Your statement "that would be the diameter" is actually a simple example of this - it's actually the radius that goes into pi*r^2, and using diameter forces you to add a constant to adjust for the fact that the diameter is double the radius! This can be extended to any other dimension anywhere else on the object, provided the proportions remain the same between model and real life. $\endgroup$ Nov 23, 2018 at 10:40
  • $\begingroup$ the issue is is this a long thin airship or a short fat one ie the length is not always the same ratio to the diameter as different airships are different shapes : the ratio holds of course, between real and model... $\endgroup$
    – Solar Mike
    Nov 23, 2018 at 10:42
  • 1
    $\begingroup$ This question is specifically regarding dimensional analysis - the dimensions (drag depends on constant*Length^2) will be the same for any shaped airship. You just need to have the same shaped model so that your experimentally measured constant applies to both model and actual. $\endgroup$ Nov 23, 2018 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.