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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.

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  • $\begingroup$ L will be a length... will it be the diameter or the length from nose to tail? $\endgroup$ – Solar Mike Nov 22 '18 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 '18 at 21:20
  • $\begingroup$ Length as in distance between two points. $\endgroup$ – Solar Mike Nov 22 '18 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 '18 at 1:34
  • $\begingroup$ Depending on the age of the pilot, their nose might actually grow during the course of a flight... $\endgroup$ – Jonathan R Swift Nov 27 '18 at 23:57
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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.

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  • $\begingroup$ Does the overall length of the ship always control the frontal area? That would be the diameter? $\endgroup$ – Solar Mike Nov 23 '18 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 $\endgroup$ – Jonathan R Swift Nov 23 '18 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. $\endgroup$ – Jonathan R Swift Nov 23 '18 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 '18 at 10:42
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    $\begingroup$ 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. $\endgroup$ – Jonathan R Swift Nov 23 '18 at 11:41

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