Deflection of a scaled down aircraft wing

AI Thread Summary
The discussion revolves around calculating the theoretical maximum deflection of a scaled-down glider wing made of balsa wood, with a length of 0.9m compared to the original 7.5m aluminum wing. The participants are trying to determine how to appropriately scale down the uniformly distributed load (UDL) from 30kg/m^2 for the larger wing to the smaller model. The second moment of inertia for the scaled wing is questioned, with suggestions to verify its accuracy since the stated value appears excessively large. Additionally, the impact of Reynolds number on performance is highlighted, noting that the smaller wing will operate at lower speeds and different loading conditions. The calculations for deflection and necessary experimental mass are presented, but the results raise concerns about their feasibility.
SJ1234
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Homework Statement


As a project, I have had to design and build a scaled down version of a glider wing. The actual glider wing would be made of aluminium and have a length of 7.5m and have a uniformly distributed load of 30kg/m^2, the scaled down version is 0.9m and made of balsa wood. I want to work out the theoretical maximum deflection of my scaled down wing and then test it to compare results.
For balsa wood E=16GPa, for aluminium E= 69GPa.
second moment of inertia for the scaled wing is 5.72×10-10m^4

Homework Equations


dmax=(UDLxL^4)/(8EI)

The Attempt at a Solution


dmax=(UDLx0.9^4)/(8x16x5.72×10-10)

dmax=(30x7.5^4)/(8x69xI)

How do I scale down the UDL?
 
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SJ1234 said:

Homework Statement


As a project, I have had to design and build a scaled down version of a glider wing. The actual glider wing would be made of aluminium and have a length of 7.5m and have a uniformly distributed load of 30kg/m^2, the scaled down version is 0.9m and made of balsa wood. I want to work out the theoretical maximum deflection of my scaled down wing and then test it to compare results.
For balsa wood E=16GPa, for aluminium E= 69GPa.
second moment of inertia for the scaled wing is 0.572m^4

Homework Equations


dmax=(UDLxL^4)/(8EI)

The Attempt at a Solution


dmax=(UDLx0.9^4)/(8x16x0.572)

dmax=(30x7.5^4)/(8x69xI)

How do I scale down the UDL?
You might want to double check the second moment of area for the scaled-down wing. 0.572 m4 is pretty large. That's the second moment of area of a solid square cross section which measures 1.62 m on a side. Make sure that you haven't omitted a ×10-something
 
Changed, thanks
 
It would help if you can the "polars" of both wings noting the difference in Reynolds number in both cases. XFOIL is a program that can calculate polars if you can't find existing data. The scaled down version will have a smaller wing chord and travel at slower speed, so the Reynolds number will be less. The speed will be related to the wing loading more than the scale factor. If the scale ratio is 1/r, then to get the speed to scale down by 1/r, the mass of the smaller model would need to be (1/r)4 of the full size glider. More on this at this web site;

http://www.charlesriverrc.org/articles/design/ibtherkelsen_scalespeed.htm
 
The scaled down wing has a length 8.3 time smaller than the original, so the UDL of 30kg/m^2 would become 30/8.3^4 6.3x10^-3kg/m^2
giving: dmax=(6.3x10^-3x0.9^4)/(8x16x5.72x10^-10) = 56455.283m. (Obviously too big)
The area of the scaled down wing is 0.08m^2, therefore the mass I need to apply in experiment is: (0.08)(6.3x10^-3)=5.04x10^-4kg
(This seems a lot too small)
 

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