ximian

10-25-2009, 12:17 PM

So I did a quick shoot of my serve from four different angles while trying out something slightly new in my motion. On a couple of my serves, you can calculate how many frames the ball takes from racket to court, so I decided to calculate my serve speed. Here's how it went down:

Here's my serve:

http://www.youtube.com/watch?v=VAU7bbivtDA

On the first serve, I counted 12 frames from the point of impact to touching the service line. The camera shoots at 30 fps, so it took .4 seconds to travel that distance.

dt = .4 seconds

Next, I need to calculate what that distance was. The ball hit the service line, almost halfway (about 40%) between the T and out wide. Using the dimensions of a tennis court:

Length from baseline to service line: 18.29 meters

Length from T to out wide: 4.115 meters

40 % of T to out wide: 1.646

Next, I need to somehow figure out how high the point of impact was. On some of the side serves you can tell that I'm about 6" off the ground when I hit, but I'm hunched over. I took some quick measurements and roughly calculated that I'm about 9' up. Or 2.74 meters.

Now I can use Pythagoras and find that the total distance the ball traveled is

sqrt[(18.29^2)+(1.646^2)+(2.74^2)] = 18.57 meters

d=18.57 m

I know that these measurements are approximate, because I don't know the exact height of impact, I'm a little inside and to the left of the court when I strike the ball, and I'm making my best guess as to where the ball lands. But the fact is, a margin of error of .5 meters still won't make a big difference on the overall speed, so it's still going to be a pretty accurate measurement.

Next, I need to account for drag. Think about balling your hand up and sticking it out the window of a car moving at 100 mph, and you'll get a similar idea of what sort of force drag is putting on my serve. That's pretty significant.

The formula for drag, if the object is between a few mm and a few m, and it's traveling under 200 m/s near the earth's surface is:

D=(1/4)(A)(v^2)

where A is the cross section, and v is it's velocity. The cross section of a tennis ball is simply (pi)(r^2). So I need the radius of a tennis ball, which is roughly:

r=0.0333 m

The problem with drag is that I need to know it's velocity, which is what I'm trying to calculate in the first place. However, since this is going to be a rough estimate of the type of force drag will place on a tennis ball, I can calculate the velocity of the ball using just distance and time, add a few mph for good measure, and use that in this calculation. So velocity is distance traveled over time, or:

v= d/dt

v= 18.57/.4

v= 46.425 m/s or 103.9 mph

Add a few m/s for a rough factor of drag, and I'll use 50 m/s (111.8 mph) as the velocity in my Drag equation. So:

D = (1/4)(pi)(r^2)(v^2)

D = 2.18

I'll leave the units of D out since that's a little complex. But I do know that for the drag equation to work, you need all your units in kg, m, and s, which I did.

Ok, so now that I have my force from Drag, I can take another look at the flight of this tennis ball.

If you draw a force diagram (a picture of all the forces acting on the ball) of the tennis ball right after it's struck, you'll notice just two forces (there are others, but they're negligible):

Gravity, or F(g)

Drag, or D

But the flight of the ball is not perfectly horizontal. It's heading down from a height of 2.74 m, so the axis is slightly tilted. Note that F(g) is pointing straight down, while D will oppose the motion of the ball, traveling slightly down. Thus, using Newton's second law (F=ma) and since we are only interested in the forces that oppose motion (x-axis), then we can calculate all the forces in the x-axis find the ball's acceleration. We expect it to be negative because the ball is slowing down due to drag.

F(g) + D = ma

where:

F(g) = mg

D = -2.18 (negative because it opposes motion)

m = 0.0567 kg (mass of tennis ball)

g = 9.8 m/s^2

a = acceleration

But let's take another look at gravity. Gravity is pointing down, so most of it's force is in the y-direction. But since the axis is tilted, there will be a small amount of gravity that is actually speeding the ball up. This component is equal to sin(theta) where (theta) is the angle measured from the court up to the point of impact.

(theta) = sin^(-1) [2.74/18.57]

(theta) = 8.5 degrees

So the final equation for acceleration is:

a = [sin(theta)mg+D]/m

a = -37.00 m/s^2

This seems a bit high doesn't it? But think back to the window and the car. Sticking your hand out the window and you'll definitely feel a strong force. That force is big enough to cause a deceleration like this. Plus, as we'll see in a bit, this acceleration force ends up looking much more rational when it's plugged into the proper equation.

So now that we have acceleration, distance, and time, we can finally calculate it's initial velocity. Newton's laws give rise to Kinematic equations, one of which is:

P(final) = P(initial)+(v)(dt^2)+(1/2)(a)(dt^2)

We have all the data except v, so a quick number crunch gives:

18.57 m = 0 + v (.4^2) - 2.96 m

The acceleration term calculates to just under 3 meters. Or in other words, the force of drag means that the ball traveled 3 meters less than it would have without drag. This seems reasonable. Solving for v gives:

v = 53.825 m/s or 120.4 mph

But let's look at our assumptions. I assumed that the height was 9' when I struck the ball, and that dt is .4 seconds, and that the mass and the radius were the numbers given (found from this website: http://hypertextbook.com/facts/2000/ShefiuAzeez.shtml). These are all reasonable, but just to be safe let's take .5 m off the distance traveled and recalculate. In that case, it's still 52.575 m/s or 117.6 mph. Still in the same ballpark. The other terms are relatively negligible as well, so I'll conservatively say that first serve was probably at least 118 mph, possibly as high as 121.

Additionally, I can calculate the difference between initial velocity and final velocity, using this equation:

v(final) = v(initial)+(a)(dt)

using a conservative 118 mph (52.57 m/s) as my initial velocity, I get:

v(final) = 37.77 m/s or 84.5 mph

This seems about right. The ball is certainly not traveling even 85 mph when the returner hits the ball, and we know that the impact between the ball and the court will slow it down more. These results look right.

I must admit I'm a little surprised it measured that fast. I was expecting something around 110, considering I wasn't trying to hit it really hard, and the tennis balls were pretty flat... But it is what it is. Now I'm curious what I can measure with new strings / balls / and trying to for a really big one!

Thoughts?

Here's my serve:

http://www.youtube.com/watch?v=VAU7bbivtDA

On the first serve, I counted 12 frames from the point of impact to touching the service line. The camera shoots at 30 fps, so it took .4 seconds to travel that distance.

dt = .4 seconds

Next, I need to calculate what that distance was. The ball hit the service line, almost halfway (about 40%) between the T and out wide. Using the dimensions of a tennis court:

Length from baseline to service line: 18.29 meters

Length from T to out wide: 4.115 meters

40 % of T to out wide: 1.646

Next, I need to somehow figure out how high the point of impact was. On some of the side serves you can tell that I'm about 6" off the ground when I hit, but I'm hunched over. I took some quick measurements and roughly calculated that I'm about 9' up. Or 2.74 meters.

Now I can use Pythagoras and find that the total distance the ball traveled is

sqrt[(18.29^2)+(1.646^2)+(2.74^2)] = 18.57 meters

d=18.57 m

I know that these measurements are approximate, because I don't know the exact height of impact, I'm a little inside and to the left of the court when I strike the ball, and I'm making my best guess as to where the ball lands. But the fact is, a margin of error of .5 meters still won't make a big difference on the overall speed, so it's still going to be a pretty accurate measurement.

Next, I need to account for drag. Think about balling your hand up and sticking it out the window of a car moving at 100 mph, and you'll get a similar idea of what sort of force drag is putting on my serve. That's pretty significant.

The formula for drag, if the object is between a few mm and a few m, and it's traveling under 200 m/s near the earth's surface is:

D=(1/4)(A)(v^2)

where A is the cross section, and v is it's velocity. The cross section of a tennis ball is simply (pi)(r^2). So I need the radius of a tennis ball, which is roughly:

r=0.0333 m

The problem with drag is that I need to know it's velocity, which is what I'm trying to calculate in the first place. However, since this is going to be a rough estimate of the type of force drag will place on a tennis ball, I can calculate the velocity of the ball using just distance and time, add a few mph for good measure, and use that in this calculation. So velocity is distance traveled over time, or:

v= d/dt

v= 18.57/.4

v= 46.425 m/s or 103.9 mph

Add a few m/s for a rough factor of drag, and I'll use 50 m/s (111.8 mph) as the velocity in my Drag equation. So:

D = (1/4)(pi)(r^2)(v^2)

D = 2.18

I'll leave the units of D out since that's a little complex. But I do know that for the drag equation to work, you need all your units in kg, m, and s, which I did.

Ok, so now that I have my force from Drag, I can take another look at the flight of this tennis ball.

If you draw a force diagram (a picture of all the forces acting on the ball) of the tennis ball right after it's struck, you'll notice just two forces (there are others, but they're negligible):

Gravity, or F(g)

Drag, or D

But the flight of the ball is not perfectly horizontal. It's heading down from a height of 2.74 m, so the axis is slightly tilted. Note that F(g) is pointing straight down, while D will oppose the motion of the ball, traveling slightly down. Thus, using Newton's second law (F=ma) and since we are only interested in the forces that oppose motion (x-axis), then we can calculate all the forces in the x-axis find the ball's acceleration. We expect it to be negative because the ball is slowing down due to drag.

F(g) + D = ma

where:

F(g) = mg

D = -2.18 (negative because it opposes motion)

m = 0.0567 kg (mass of tennis ball)

g = 9.8 m/s^2

a = acceleration

But let's take another look at gravity. Gravity is pointing down, so most of it's force is in the y-direction. But since the axis is tilted, there will be a small amount of gravity that is actually speeding the ball up. This component is equal to sin(theta) where (theta) is the angle measured from the court up to the point of impact.

(theta) = sin^(-1) [2.74/18.57]

(theta) = 8.5 degrees

So the final equation for acceleration is:

a = [sin(theta)mg+D]/m

a = -37.00 m/s^2

This seems a bit high doesn't it? But think back to the window and the car. Sticking your hand out the window and you'll definitely feel a strong force. That force is big enough to cause a deceleration like this. Plus, as we'll see in a bit, this acceleration force ends up looking much more rational when it's plugged into the proper equation.

So now that we have acceleration, distance, and time, we can finally calculate it's initial velocity. Newton's laws give rise to Kinematic equations, one of which is:

P(final) = P(initial)+(v)(dt^2)+(1/2)(a)(dt^2)

We have all the data except v, so a quick number crunch gives:

18.57 m = 0 + v (.4^2) - 2.96 m

The acceleration term calculates to just under 3 meters. Or in other words, the force of drag means that the ball traveled 3 meters less than it would have without drag. This seems reasonable. Solving for v gives:

v = 53.825 m/s or 120.4 mph

But let's look at our assumptions. I assumed that the height was 9' when I struck the ball, and that dt is .4 seconds, and that the mass and the radius were the numbers given (found from this website: http://hypertextbook.com/facts/2000/ShefiuAzeez.shtml). These are all reasonable, but just to be safe let's take .5 m off the distance traveled and recalculate. In that case, it's still 52.575 m/s or 117.6 mph. Still in the same ballpark. The other terms are relatively negligible as well, so I'll conservatively say that first serve was probably at least 118 mph, possibly as high as 121.

Additionally, I can calculate the difference between initial velocity and final velocity, using this equation:

v(final) = v(initial)+(a)(dt)

using a conservative 118 mph (52.57 m/s) as my initial velocity, I get:

v(final) = 37.77 m/s or 84.5 mph

This seems about right. The ball is certainly not traveling even 85 mph when the returner hits the ball, and we know that the impact between the ball and the court will slow it down more. These results look right.

I must admit I'm a little surprised it measured that fast. I was expecting something around 110, considering I wasn't trying to hit it really hard, and the tennis balls were pretty flat... But it is what it is. Now I'm curious what I can measure with new strings / balls / and trying to for a really big one!

Thoughts?