I guess the newly design shuttle without following the tradition flight behavior would not be badminton shuttle anymore.... =P
em yea. high school student got lost after the first set of calculations. jeez you people need to get a dam life. yea im a badminton fantaic to but trying to figure this out is near impossible. just go ask the Yonex manfuactuers. jeez be smart and call the people. oww my aching head. the numbers!!!!!!!!!
Something for me... Interesting topic. I was trying to write the basic equations (the Cx one) and I found that I forgot it. It is 8 years when I left university. I had to deduce. Am I correct that the equation is F=Cx*S*v*v where S is area in axe x cut, v is velocity and Cx is coefficient of air resistance. If I remember correctly this equation is valid untill the turbulence appeares. I assumed Cx as 0,3. It is totally wild guess. The velocity is then 22m/s. If the Cx is 0,15 (I seriously doubt - too little) the velocity is then 15m/s. If Cx is 0,6 (again I seriously doubt - too much) the velocity is 30m/s. I am quite confident that real Cx will be in interval 0,15-0,6. Conclusion - the velocity you are seeking should be in interval 15-30m/s. In the normal game you will always hit the shuttle falling with positive acceleration towards earth. Please do not stone me. I know that this was more quessing than facts. By the way in shuttlecock case the Cx is nonlinear function of velocity which makes any precise calculation more less impossible. The only right approach is the first suggested by Kwun. We have to find it empirically.
Correction of results. I overlooked one zero. Cx 0.3 - velocity is 7m/s. Cx 0.15 - velocity 5m/s, Cx 0.6 - velocity 10m/s. So the conclusion was wrong. Terminal velocity for shuttlecock is somewhere between 5 and 10 m/s. Sorry
Erm so what you're saying is that if Drag coefficient is higher the terminal velocity is higher? Hunh?
Oh well, the summer is too hot in here... I did not calculate it from formula, just multiplyed by square root of 2 without thinking. You are 100% right - it should be 10m/s for Cx = 0.15 and 5m/s for Cx = 0.6. Shame on me! Call me miss bikini if you like.
This thread is trap. Finally I found my old textbook. I was wrong in some statements. The correct formula is F=1/2*ro*Cx*S*v*v where ro is density of the gas. The formula is valid for turbulent flow not for laminar flow. Hence last correction of results Cx=0.15 - v=7m/s, Cx=0.6 - v=14m/s.
hey trapped, why do u want to calculate the Vt of the shuttle when it had been experimentally measured and calculated already? Second, there isn't a fix drag coefficient for the shuttle because shuttle spins along its path and deforms according to velocity. Also, S is hard to determine as a shuttle is not a rigid body except for the cork. Your lower range assumption of Cx is too low. Cx of 0.15 is more streamline than the most experimental streamline cars ever built. Streamlined body 0.1 Sports Car 0.2 - 0.3 Sphere 0.47 Typical Car 0.5 Station Wagon 0.6 Cylinder 0.7-1.3 Racing Cyclist 0.9 Truck 0.8 - 1.0 Motorcyclist 1.8
Easy answers... Ok, this thread is quite unlucky for me. You probably noticed how many stupid mistakes I have commited. Sorry for that. But the last result is final. I have answers to all of your questions: I was not aware that Vt was already measured. I understood the thread as challenge to find it. But even if it is measured I would calculate it for pure joy. I mentioned earlier in this thread that there is not a fixed drag coefficient. Try to read my contribution once more. S is actually very easy to measure. It is area of axe X cut (2D). I am not sure if I use correct english terminology. In case I do not please relate to Newton formula where it is described what S means. And finally - you spectacullarily failed to understand the basic idea of the calculation. And I spectacullarily failed to explain it in such a way that everyone can understand. Cx 0.15 was the lower range of possible drag coefficient. Cx 0.6 was the upper range. I expect the real Cx to be somewhere in between. In the same time I expect the calculated Vt to be somewhere between 7 and 14 m/s.
i'm not trying ruin your fun in calculating the Vt but your approach and result were too simplistic and inconsistent. Let us assume that your use of equation F=CxSv^2 was true where Cx =0.15 and your calculated Vt=7m/s and where Cx = 0.6 and your calculated Vt=14m/s where Cx of 0.15 is your estimated lower range of drag coefficient and Cx of 0.6 is your upper range. (here, your result is invalid already, why does Vt increases with drag coefficient Cx??) Since the measured and published Vt was 6.8 m/s, using your logic, it means the Cx has to be less than 0.15. Cx at this range is representative of a very extremely aerodynamic object, not logically representative of a feather shuttle. Take the figure from the table i posted, Cx of a motorcycle is 1.8 and if u drop a motorcycle from the sky, i guess the Vt is around 140 mph (62.6 m/s). How can you say an object with a Cx of 1.8 can have a Vt about 9 times of a shuttle having a Cx less than 0.15 (from your calculation of above)??
S is actually very easy to measure. It is area of axe X cut (2D). I am not sure if I use correct english terminology. In case I do not please relate to Newton formula where it is described what S means. S for feathers are NOT easy to determine. Feather is not a smooth surface object, it is comprise of many interconnected small and microscopic vanes, they move and flex according to air stream moving across them. Take a look at the shuttle along its longitudinal axis, is has holes and passages between vanes where air can pass through. I hope you don't mean using a conial formula or shuttle radius to calculate the surface area.
I am really not in a mood nor have time for going into useless and endless arguing. If the measured velocity is 6.8m/s I was quite close to reality with my oversimplyfied model. And as I said earlier for meaning of S relate to description of Newton formula. By the way it is F=1/2*Ro*Cx*v*v.
Yes, your guess was quite close but it's was on the opposite extremes and the ranges were quite large (Cx between 0.15 and 0.6 is like between a bullet and a station wagon; Vt between 7 to 14 m/s or 15.65 to 31.3 mph, i would get a speeding ticket if i'm 6.2 mph over speed limit here). However, not to ruin your fun, i have taken the simplified newton equation and try to see how it pans out. Using known parameters, the resulting expression is (i prefer to use Cd instead Cx, and A instead of S) Cd A = .00173 for a 5 g shuttle from here, i have to make guesses because we really don't know the effective area or the real Cd at the Vt condition. Assuming the shuttle is within the mid range of IBF approved diameter (63mm) and 10% slippage of air through the holes and vanes, I got a Cd of an average shuttle of ... 0.616 To highlight what i've said above about opposite extremes, you have shown that: Cx=0.15, Vt=7 m/s and Cx=0.6, Vt=14 m/s a closer to reality is Cd=0.616, Vt = 6.8 m/s ie, you were close to your assumed lower Vt of 7 m/s while the realistic Cd came close to your upper range of Cx of 0.6. You have really pushed your guessing ability
