http://www.donthireddy.us/tennis/speed.html
PS: Thanks Mike, for posting the correct link while I have been AWOL.
PS: Thanks Mike, for posting the correct link while I have been AWOL.
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Mike Cottrill said:http://scienceworld.wolfram.com/physics/DragCoefficient.html shows a hard hit tennis ball with a drag of 0.3 and using Sampras’ serve as a reference you get ~.003. Okay physics guys, prove this formula acceptable.
Mike
maverick1 said:Sorry, how did you get the 0.003?
I don't see how you can do anything useful with the The coefficient as defined at Mathworld, without knowing what the "size scale" L is. Presumably it would some function of diameter of the Tennis ball.
The coefficient "c" I used above is equivalent to 0.5 * Cd * Rho * L^2 / m
where
Cd - Mathworld's coefficient of drag
Rho - density
L - Size scale
m - mass
Since the mathworld page says nothing about L, this info is useless.
drakulie said:Guys, how do I count the frames? I've tried with no luck. Any suggestions?
Sorry, I forgot you used a digital camera, not a camcorder. My previous advice is useless.drakulie said:Guys, how do I count the frames? I've tried with no luck. Any suggestions?
I just got an ice cream headache.maverick1 said:This is a spinoff from another thread:
http://tt.tennis-warehouse.com/showthread.php?t=120776
This is what I said there:
----------------------------------------------------
If your own serve took t frames in standard 30 fps video to cover a distance of d feet in the air,
your serve speed in mph was:
(exp( d / 5280 * 28.3) - 1 ) / 28.3 / (t / 108000)
To make it easier, here is a table converting from the number of frames of standard 30 fps video it took for a down the middle serve to land on the T.
Frames Speed
11 -> 131.6
12 -> 120.6
13 -> 115.8
14 -> 103.4
15 -> 96.5
16 -> 90.5
17 -> 85.2
--------------------------------------------------
I think it is correct, but I went thru a lot of pretentious math that I normally don't do. For all I know it may be unnecessary. I would be happy to hear any comments from the science grad students who I know prowl this forum.
Here is the derivation ....
The model for motion under a drag force is
dv/dt = - k * v^2 /m
Note that this is a statement of Newton's second law,
acceleration = Force/mass.
Drag force is -k*v^2 where k is
a constant related to the drag coefficient.
Let us make up a modified drag cofficient c = k/m to simplify our job
since mass will always be the same(that of a tennis ball)
dv/dt = -c*v^2
The solution to the above separable first order differential equantion is
v = 1/(c*t + c1)
c1 is the constant of integration, and when you set t=0 above, you can see that
c1 must be equal to reciprocal of initial velocity v0
so,
v = 1 / (c*t + (1/v0)) -------> Equation 1
Integrating the solution for v, we get distance
s = ln(c * t + (1/v0))/c + c2
c2 is yet another constant of integration.
Knowing that the distance s is 0 when time t=0,
c2 = -ln(1/v0)/c
The formula for distance is
s = ln(c * t + (1/v0))/c - ln(1/v0)/c
==>
s = [ ln(c * t + (1/v0)) - ln(1/v0) ] / c
==>
s = [(ln(c * t + (1/v0)) / (1/v0) ) ] / c
==>
s = [ ln(v0 * c * t + 1)) ] / c --------> Equation 2
==>
exp(s * c) = v0*c*t + 1
==>
v0 = [exp(s * c) - 1] / (c * t) ------> Equation 3
This v0, the initial velocity is our goal. From our camcorder
experiment, We know s, the distance
traveled and we know t, the time it took to travel that distance.
The only unknown in the above formula is c, the modified drag
coefficient for a Tennis ball. Time to claibrate our model
using a known fact about Sampras' serve - that a 120 mph serve slows
down to 87 mph over 60 feet.
( http://wings.avkids.com/Tennis/Project/speed-02.html)
Substituting the numbers in Equation 1,
87 = 1/(c*t1 + (1/120))
t1 above is whatever time that 120 mph took to travel 60 ft(= 60/5280 miles).
c*t1 = 1/87 - 1/120 = 0.00316092 ----> Equation 4
Now using Equation 2,
60/5280 = [ ln(120 * c * t1 + 1) ] / c
We can plug in the expression for c*t1 from Equation 4
60/5280 = [ ln(120 * 0.00316092 + 1) ] / c
c = [ ln(120 * 0.00316092 + 1) ] / (60/5280)
c = 28.3
We plug this c into Equation 3 to get
v0 = [exp(s * 28.2) - 1] / (28.3 * t)
We want to be able to express time as number of video frames.
t = (F / frame_rate_sec) / 3600
The 3600 is to convert time to hours. If distance s is expressed in feet want to convert the
distance s to miles
v0 = [exp(s / 5280 * 28.2) - 1] / (28.3 * (F / (frame_rate * 3600 )))
After eliminating some parentheses, We arrive at the final
formula for Serve Speed
==================================================================
v0 = [exp(s / 5280 * 28.2) - 1] / 28.3 / F * frame_rate * 3600
==================================================================
where
v0 is the initial speed of the serve,
s is the distance traveled by the serve duting the obeserved interval
F is the number frames of video it took the ball to travel this
distance
frame_rate is usually 30(I think 29.97 is the precise value) with most
camcorders.
why, if he knows the frame rate? Which in most cases is 30 or 15fps.maverick1 said:Sorry, I forgot you used a digital camera, not a camcorder. My previous advice is useless.
norcal said:I just got an ice cream headache.
Mike Cottrill said:why, if he knows the frame rate? Which in most cases is 30 or 15fps.
maverick1 said:Because I assumed digital cameras(as opposed to camcorders primarily meant for video) had no sophisticated playback features. I have never used a camera for video.
I don't know much about these things and I will stay out of it now.
Mike Cottrill said:I have run the number through some of my videos and it looks pretty darn close. Good job. Now we the physics guys to give the formula the final approval.
drakulie said:On the 101 mph one, it is a little more clear when the racquet makes contact with the ball and is reaching the ground exactly on the 14th frame. When I reach the 15th frame, it is already bouncing, more so than the 100 mph one.
I NEED SOME MONEYYYYYYEYEYEYAYYEYEYEYYEYEYEYEYEYEYEYEYEYYYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYEYYEYEYEYEYEYEYEYEYEYYEYEYEYTen.Is said:Hmmmmmmmm...LETS PURCHASE A RADAR!!!!!!!!!!!!!!!!
For sme reason, A lot of sites have trouble with the host name mavericks.ccZets147 said:Link is down I assume.
Yes, this was my first attempt at writing JavaScript. Was surprisingly easy. Just googled for an example, found a mortgage calculator and modified it.Mike Cottrill said:Very nice. I did a c program, but HTM allows all to do the calculation.
We need to make this a sticky. Most likely in the Tips section.
Mike
drakulie said:here are some serves that were recorded as the ball approached the baseline.
this one is 79 mph:
http://www.youtube.com/watch?v=Uo4bkssVVY4
this one is 82 mph:
http://www.youtube.com/watch?v=k8JkyTBiyPk
this one was recorded at the back fence (21'3" away from baseline) at 43 mph:
http://www.youtube.com/watch?v=LKLUEvGwIes
drakulie said:It doesn't necessarily specify an exact range, so I am not sure. I am aware the ball speed has been recorded before it reaches the baseline, I just thought it was interestsing to see the speed as it approaches the baseline and back fence.
As for the picture, I can't get it any larger. On my computer it is much larger and you could clearly see the ball landing. It is directly inside the service line and on the T. For some reason the picture shrinks when I transfer it.
By the way, the video is a ".mov" extension when it is downloaded from my camera to the Kodak software.
drakulie said:mav and mike, here is a video of 101 mph down the T.
http://www.youtube.com/watch?v=5EJe64Ky7rE
here is the action shot of the video, frame by frame. The first frame is "contact" and at the 14th frame the ball has struck the ground. The 15th and 16th frame the ball has already left the ground.
I didn't. drakulie emailed the original to me and I put it up on my website.ShooterMcMarco said:how did you enlarge those pics?
it seems that this method of calculating speed completely ignores the spin of the ball which would lead to a different drag force. Neglecting ball spin introduces inaccuracy 1-10 percent depending on the type of serve. I would suggest rotational speed is 10% of linear speed on first and 30% on second based on http://wings.avkids.com/Tennis/Project/usspin-07.html
At which point do they measure serve speed at the professions tournaments anyway?
about using the video to find the speed
lets say a camera records @ 30fps
it doesnt mean that each frame is taken 1/30th of a second after another ?
:S
lol okay but that with better cameras (for tv broadcasting)vincent, this is irrelevent. Note that they compared using this method with radar results at the australian open and found that the formula provided reasonably accurate results.
Hi everyone.
I understand the formula to an extent, but that link is no longer a calcualtor (It looks like a list of names to me).
If it is decent quality and the video is in the right format it can be done. Film from behind and have the camera were you can see the ball contact the racquet and impact with the court.If I were to film my serve, could someone tell me how fast it was going?
http://www.donthireddy.us/tennis/speed.html
Acording this, The calculation below relies on the reported finding that a serve with an initial speed of 120 mph slows down to 87 mph over a distance of 60 feet
So we have a 27,5 % of variation of decreasing speed in 60 ft.
then in 36 ft(more or less the distance between server and net , i mean from where the ball is hit to the net(where radar guns are normally located)): we got some 16, 5% of variation in every serve.
So in my case i got a spped track xtreme radar gun set at net, and i got 106 mph at net, but the real speed(ATP SPEED, WHEN THE BALL IS HIT), i mean the atp/ initial speed will be more or less a 16- 17 % faster. So 106 mph at net means i actually have 123, 3 mph of initial speed.
What do you think guys? am i wrong?