interesting idea coming from jwu's comment: the higher the shuttle travels, the faster it will come back down because of our wonderful friend, gravity as there is two force in play here, gravity and air resistance, has anybody wondered what the terminal velocity of a shuttlecock is? terminal velocity is when the object falls to a point in which the air resistance and gravity are at equilibrium and that's the fastest an object will fall....

well, I doubt any of us can experiment with this concept in any gyms we play at. c'mon all your physicist out there, time to crunch some numbers.

i was actually hoping someone who lives in high-rise building would take a camcorder outside, and then have someone drop a shuttle from say the 5th-10th floor. and then like what we did for strobe photography in high school, do a frame by frame measure of the distance travelled. we should see it reach terminal velocity.. someone with a lot of free time in their hands...you know, those students among us who are heading into their summer vacation....

i was practicing my wrist lift on a high ceiling court (50 ft). It looks like the shuttle comes down at around 12 to 18 mph. This is a visual estimate

Interesting question. It's been 10 years since I had physics, but I seem to remember terminal velocity is standardized at about 32 m/s (is that right?). However, the falling shuttle should be much slower because of 1) the shape of the bird offers more wind resistance, and 2) nobody hits it so high that it can accelerate to that velocity (gravity is 9.81 m/s^2, so it would need to fall more than 3 seconds to reach 32 m/s). I don't think a test off a building would be accurate either since there can be significant updraft.

hehe, I would love to drop a shuttle out of a plane and watch the baby fall 10,000 feet lol...i have no idea why that is funny. Matt

You can determine the terminal velocity by clearing the shuttle upwards and watching it drop. A shuttle should reach terminal velocity almost right away. There is no set terminal velocity... it's determined by a number of factors such as medium density (in our case atmospheric pressure) and (air) resistence. Given the high air resistence and substantial medium density, the terminal velocity is likely to be very low. As for jwu's comment, it's is true only before the shuttle reaches terminal velocity of course. (It's final speed will be the same if you drop it from 20 feet versus 2,000,000 feet under the same conditions.)

Actually BRL is possibly right ... if the shuttle reaches terminal velocity in under 20 ft then BRL is right, if not, then he is wrong. It wouldn't really be that hard to work out with minimal error if you had two ppl and a reasonably high ceiling (>10 ft or 3m)*. Basically you measure the height that you drop it from. 1. Drop the shuttle, cork facing downwards, 2. Measure how long it took to hit the ground, 3. Repeat 1 and 2 a few times and average the result. 4. From that you can work out the acceleration. 5. From that you can work out the drag coefficient. 6. From that you can work out the terminal velocity. To the guy that somehow remembered that terminal velocity is standardised at 32 m/s, I'm not sending my kids to your school. I can't see how anyone would try to teach you that there is a standard terminal velocity I mean drop a 20 cm radius steel ball bearing (big ball bearing I know ) from a plane, and the same ball bearing tied to a working parachute and see if they attain the same max speed NB for most scenarios, Terminal Velocity is attained when: the force due to drag is equal to the force due to gravity. NB I said force, not acceleration. In this case with feathers, buoyancy may be significant, therefore Terminal Velocity will be attained when: Drag Force + Buoyancy Force = Gravitational Force. Have fun. *Funnily enough this will only work if the shuttle does not attain terminal velocity before it hits the ground, I suggest you don't try this on really high ceilings, 10-15 ft is probably ideal since you can easily hit a shuttle that high and I'm pretty sure it hasn't attained terminal velocity by the time it hits the ground.

Hehe, I guess 10 years is a long time. Although I do understand the variables involved, I'm not sure why I got the number 32 m/s stuck in my head.. I thought it was some number from an idealized situation, but I'm obviously wrong. Oh well, we all make mistakes. Thesis writing is turning my brain into jello.

I have no idea which part of my reponse you are referring to, so i'll just respond to that equation you posted. wrong, G=f(r) technically means earth's gravitation constant (G) is a function of the distance from earth of an object (r). Since it's a constant by definition it cannot be the answer to a function. So i'll presume you meant g=f(r) If you mean gravitational acceleration is a function of the distance from the surface, then the difference would be so small that it's irrelevant. Besides, the difference would be between locations (i.e. toronto versus calgry) rather than between heights (i.e. floor v roof) at the heights we are dealing with. Since the shuttle is within the atmosphere of earth and we are on planet earth, that function has no use to us. It's not even a complete forumla.

yes, that wasnt a formula but a statement in short form, and yes, i meant g force and not earth gravitational constant. I made that statement because of your example of 20 vs 2e6 ft.

true. environmental condition determines the density of air, which affect air resistance, which affects terminal velocity. however, as badminton fanatics, the environmental condition we are interested in is of course the "standard", ie. when we hit a shuttle underhand from baseline, it should land around the double service line... i am not sure if the birdie would reach terminal velocity in 30 or so feet. that's what i am trying to determine here. we can guess, but we need to somehow verify. terminal velocity is reach when the force of gravity equals the force of air resistance. F_g = F_r as we know, F=ma, m = 5g or 0.005kg, a = 9.8m/s^2, thus: F_g = 0.049N at what speed does the shuttle need to travel to create an air resistance of 0.049N?

kwun, which axis are interested in the terminal velocity, x (horizontal) or y (vertical)? If both, then the x component is dependent on how hard the shuttle is strike initially and its angle. For rough estimation, distance to reach Vt(y) of a falling shuttle is 5, 7.5 and 11 feet for Vt(y) of 12, 15, and 18 mph

woohoo, my eye ball estimate came pretty close to my mid point of 15 mph, i've over estimated my error limits by +/- 3 mph. Experimental result was 15.2 mph. http://electron.physics.buffalo.edu/ComPhys/Chapter3/styro.pdf

Seems to me it would be hard to calculate the air resistance of a shuttle with all the variables involved. I've always wondered to what extent Yonex or any shuttle manufacturer has gone with scientific testing of shuttles, such as wind tunnel tests. There might even be a computer program now that could design a shuttle to fly at any particular speed. Of course, producing a shuttle with natural ingredients to meet those specs is another story (would be easier with all-synthetic shuttles).

but we are dealing with terminal velocity not instantaneous velocity. Even if the bird is shot out of a cannon near the speed of light from space, our atmosphere would probably slow it down enough such that by the time it hit sea-level, it would have slowed to terminal velocity--the speed as dropping it from 2,000 feet or 200 feet, etc etc. Presuming we're using a super-bird that is exactly like a regular bird doesn't burn up on re-entry.

It's actually very easy to test the air resistence if you have the right setup. If I wanted to find out I would place a spring behind the shuttle and blow on the shuttle with a fan large enough to release a uniform speed over the shuttle. From the speed of the wind and the compression of the spring, we can determine air resistence of a shuttle. Since springs give answers in force (N), then kwun did half the work already. I have fans that i know the CFMs of. By mounting it in a pipe, I can determine the wind speed from the pipe's cross-sectional area and wind volume. All I need now is a small spring to finnish the calculations. But where to buy get a spring that's sensitive enough to compress at 0.5N!!?? The smallest springs I have are 1/8inch and that is too big already.

OK, you're talking about empirically determining the air resistance of a whole shuttle. That's certainly the easy way, and the best for the purpose here. What I'm talking about is starting with the air resistance of goose feather material and the cork covering, and then using such factors as the width and length of the feathers, the cut of the ends, the angle they protrude from the cork, the angle of the feather relative to the one next to it, the radius of the cork, etc., etc. That's why I said it would be difficult, but not impossible with the kinds of engineering software we have today. Shuttle design has evolved by trial-and-error, with minor changes to optimize flight behavior. But what if you wanted to design one from scratch?