Hi, I'm new to the forums here, just wanted to know the surface area of a standard 5g Yonex shuttle in contact with the air when in descent. You know, the cone-like area on the outside, not the inside. Doing an Extended Essay on it. Anybody's help will be greatly appreciated! (Sorry if it's been answered before somewhere, search button didn't work for me, stupid IE)

I doubt if anyone has considered it before. You are mostly looking at the area of the feathers, plus a bit more for the spines and stuff. I would guess it would be of the order of 20 cm^2

wrap some tinfoil around the area and then just measure it once its been cut exactly to shape. its overly simplified as wouldnt air flow through the gaps and hit the inside of the feathered circle too? another way would be to measure a plastic shuttle, and just cut off the plastic sheeath at the cork. calculate the corks contact area(half a sphere of that diameter-(measured on the flat end of the cork) ) and then unroll the sheath and work that out too.

I hope to oversimplify it by assuming the shuttle is solid; as in there are no gaps for air to flow through. If that's the case, does anybody know how to calculate the area? Now there's the semi-sphere, which is easy to calculate, but the almost-cone-shape? How do you calculate that?

i still think wrapping paper or tinfoil around it to fit exactly , then unfolding and calculating the 2d shape is easier.

I did a similar project for maths A-Level - I'm assuming you're trying to estimate the drag coefficient. Measure the radius of the top circle Rt, where the feathers' points are (should be around 31mm), the radius of the base Rb (~13mm) and the distance between the centre of the base and the centre of the top circle of feather tips h. You can measure the length of the feathers and use trig - it's far more accurate. Then chuck them into the formula S = pi (Rb+Rt) * sqrt((Rt-Rb)^2+h^2)) (it's awful trying to type equations in here; have a look at http://mathworld.wolfram.com/SurfaceArea.html under "conical frustum". (Edit: don't forget to add the hemispherical area of the base as well: 2*pi*(Rb)^2)

Wow. Thanks a lot, I'll get started on that. On a side note, anybody here knows if drag coefficient changes as mass changes?

From my result, I ended up with the following differential equation: mg - c*v^2 = ma where c is the dynamic drag. From here, c has the dimensions "mass per length" (I hope you've come across dimensional analysis before), so c would be proportional to mass.

the real surface area of a feather shuttlecock can not be measured or known exactly. drag coef. is not mass related to the object.

I see...so it's a fixed figure; a constant. 0.616? (from http://www.badmintoncentral.com/forums/showthread.php?t=10453)