Serve toss question....

[uspta146749877]

We will wait for Don to respond

The distance a ball drops due to gravity is y=gt^^2/2, where g=9.8m/s^^2 is the acceleration due to gravity. After 0.1 seconds, y=0.049m and after 0.2s, y=0.196m.

Formula for drag force F=C A d v^^2/2, C= drag coefficient, = 0.55. A is the cross-section of the ball =pi x R^^2, R=radius of the ball. .d=density of air = 1.21 kg/m^^2. A ball experiences a back force F = 0.55 x 0.00342 x 1.21 x v^^2/2 = 0.00114 x v^^2 Newton. If v=30m/sec (180 km/hour), then F=1.026 Newton.

Force of gravity on the ball is =mg. m=0.057 kg for an average tennis ball. So 9.8 x 0.057= 0.559 Newton. Comparing this to the force of gravity on the ball: drag is twice as much as gravity.

The magnitude of the Magnus force on a tennis ball is similar to the drag force equation.
F=( 1/(2 + (v/vspin)) A d v^^2)/2 , where vspin = R x (angular speed about a horizontal axis perpendicular to the flight of the ball). Angular speed is about 16 to 80 revs per second. If v=30m/s, w about 60 revs per second, with R=0.033m, then vspin = 9.9m/s, v/vspin= 3.03. Equation shows that the Magnus force is smaller than the drag force, but can approach it the bigger the spin.

1.We will wait for Don to respond.
One may decide that a thread is closed or NOT.

A proper etiquette can be produced.

2.One can see that if vspin aproaches zero a Magnus force approaches zero

3.My main interest at THIS JUNCTURE is a slope produced by an ALMOST FLAT serve for a 4 feet forward toss.
For example whether a slope is bigger than 8 degrees or NOT.

A secondary interest is shape of of trajectories of a ball.

4.Some comparisons with Plagenhoef,Brody and Cross would be of interest as well.

5.some limitations of a Don's approach can be analyzed

Your post above (in blue) has some relation to a post below
---->
John and Don exchanged posts on related question:

http://www.tennisplayer.net/bulletin...hp?t=2&page=85

posts #845 and 846

[uspta146749877]

A drag force comparable to gravity for lower speeds

Your/Phil's quote
----
Force of gravity on the ball is =mg. m=0.057 kg for an average tennis ball. So 9.8 x 0.057= 0.559 Newton. Comparing this to the force of gravity on the ball: drag is twice as much as gravity.
--->
My response-A drag force becomes comparable to gravity when
a velocity is closer to 60 miles/hour
See that if v---> v/2
then v*v ---> v*v/4

and if v---> 2/3*v
then v*v---> 4/9 *v*v

Probably both should be taken into account-at least in some regions,for example close to a net.
A next OBVIOUS point is that a vertical component of a drag force is AN OPPOSITE to a gravity force.So a drag force does NOT bring a ball lower per se

[tennis_chiro]

Let's calm down a little

This is getting way too complicated. I was just trying to set up a way we could look at the difference between a serve hit like Soderling and one hit like Krajicek or most of today's players. I wanted to look at the window of acceptance without regard to spin or drag, just as a starting point. Now we are getting to where we need multifactorial partial differential equations. I was pretty good up through LaPlace transforms and simple stuff. I don't remember if I was ever able to handle partial differential equations. I certainly can't now.

We don't need an exact answer. We just need something that makes sense. Something the only marginally technical person might be able to understand. Now we have some equations for drag and spin (Magnus force) and we should be able to fine tune what I am trying to put forth, but first let's for just a moment put aside spin and drag. They may lift the top side of the "window" a little bit, but they certainly can't lower it. We are talking about 3" of clearance and the ball is only 2.7" in diameter. After we come to some conclusions on that basis, then we can factor additional movement of the ball due to spin and drag. I might even be able to adjust the spread sheet to do that for us. But first let's step back a little bit.

Phil, a couple of questions for you
First, if you toss the ball 4 feet into the court and you have a full reach to contact with the ball, you drop the contact point almost a full foot.

Vertical Distance up to forward toss = VDf = Square root of (reach squared - 4 foot squared)
for a 9'6" reach, VDf = SQRT[((9.5)**2) -(4)**2]= SQRT(90.25 - 16) = 8.62 ft. That is going to be below the green line on your illustration. It is significant. Why don't you want to consider that?

Two, to say gravity is a small influence is just incorrect. My calculations have the ball dropping, strictly due to gravity, almost a foot by the time it gets to the net and almost a foot and a half more before the ball hits the court at the service line. Imagine, if the great stagemanager in the sky could turn off gravity for just a moment for us, that beautiful serve you saw land right at the service line would be flying by that line 2 1/2 feet up in the air.

Also, because the vector of the path of the ball is not horizontal, it has to have a vertical component to it; granted, it is probably small, but it is directing some portion of the drag force in the vertical plane. At first glance this would actually be an upward force if it is the overall drag force vector is exactly opposite to the path of the ball. Does anyone else see that this force would be more significant near ball strike because the ball has more speed and drag is proportional to the square of the ball's velocity. Still think it would be small, but it might be a couple of inches and with this "window" being so small, that is significant. Problem is, you would need to figure out the actual angle of descent at any given instant and integrate it over the total path of the ball, first to the first barrier, the net and then to the second objective, in front of the service line.

Also, recognize that this analysis says nothing about the increased difficulty of executing the serve when you are reaching into the court at such a severe angle (I tried to measure on picture of Krajicek and he was pretty close to 45 degrees which would be even further into the court). Then there is the question of the biomechanical advantages/disadvantages to this position. You have to subjectively judge whether or not it is worth it to try to develop the extreme lean. But we could better do that if we know the effect of the forward toss on the "window".

I don't know why we are spending so much time on this. I find it interesting to play with the numbers. I think I am just procrastinating doing some of the real work I should be doing on this computer right now. I'm trying to redesign a new patient initial exam form and I've got all my paper work to do for year end with taxes. I've got Plagenhoef's and Brody's books in the mail to me now. So I will try to understand some of the more technical parts of this problem, but I think we can come to some good understanding of the advantages/disadvantages of the forward toss at a much simpler level first.

We are getting over 100 views per day on this thread. I sure wish somebody smart would speak up. Short of that let's try to make a little progress on the basic questions first.

thanks,
don

[gzhpcu]

Yes, we are getting awfully theoretical, and into muddy waters here...

Stepping back for a moment and pausing, I ask myself: Sampras had probably the best serve ever, and his ball toss was way forward and way to the left. This led to his extreme body lean, not only forward but to the left as well. Very hard to duplicate. He had so much spin on his serve, that the magnus effect played a major role. For a heavy serve, you want speed and spin, which is what Sampras achieved. A purely flat serve is not a heavy serve, and not all that difficult to get back, unless extremely well placed. So why all the theoretical interest in a flat serve?

[gzhpcu]

A next OBVIOUS point is that a vertical component of a drag force is AN OPPOSITE to a gravity force.So a drag force does NOT bring a ball lower per se

The drag force opposes the direction in which the ball is traveling due to the air density. It shortens the ball trajectory, this results the ball being lower along the trajectory.

[gzhpcu]

The 3" inch clearance over the net is only applicable in a theoretical drawing showing a straight line trajectory from the impact point to the service box. Drag and the magnus effect allow this clearance to be raised considerably. I still don't follow what a model which does not correspond to reality brings us...

[gzhpcu]

Vertical Distance up to forward toss = VDf = Square root of (reach squared - 4 foot squared)
for a 9'6" reach, VDf = SQRT[((9.5)**2) -(4)**2]= SQRT(90.25 - 16) = 8.62 ft. That is going to be below the green line on your illustration. It is significant. Why don't you want to consider that?

Here you go: the blue line... still same result...



Uploaded with ImageShack.us

[gzhpcu]

The tennis ball is special because at the speed it travels through the air, it lies around the border of being both predominantly laminar and predominately turbulent. Spin, has a gyroscopic effect, i.e. stabilizes the flight. No spin on the ball makes it more apt to turbulent flight (wobbling flight)...

For more detail, check this out:
http://people.stfx.ca/x2006/x2006nsk...nis%20bals.pdf

("Aerodynamics of spinning and non-spinning balls")

[uspta146749877]

A link above

I cannot get a link above working

[uspta146749877]

A drag force decomposition

A drag force should be decomposed into two components:
vertical and horizontal.
Assume a slope of 7 degrees.
A vertical component of a drag force will be much smaller than a gravity contribution
An original vector of 120 miles per hour will be decomposed into a large
horizontal component and a very small vertical component
A horizontal component is already in your 90 miles per our decreased speed at a net.

Please use

The drag equation calculates the force experienced by an object moving through a fluid at relatively large velocity (i.e. high Reynolds number, Re > ~1000), also called quadratic drag. The equation is attributed to Lord Rayleigh, who originally used L2 in place of A (L being some length). The force on a moving object due to a fluid is:

\mathbf{F}_D\, =\, -\tfrac12\, \rho\,A\, C_d\, (\mathbf{v}\cdot\mathbf{v})\, \frac{\mathbf{v}}{||\mathbf{v}||},

see derivation

where

\mathbf{F}_D is the force vector of drag,
ρ is the density of the fluid,[3]
\mathbf{v} is the velocity of the object relative to the fluid,
A is the reference area,
Cd is the drag coefficient (a dimensionless parameter, e.g. 0.25 to 0.45 for a car)

see

http://en.wikipedia.org/wiki/Drag_%28physics%29#Types_of_drag

[gzhpcu]

I cannot get a link above working

It is a link to a pdf document. If your browser doesn't have a pdf add-on, try this link instead:
http://docs.google.com/viewer?a=v&q=cache:lc3V-IzY-doJeople.stfx.ca/x2006/x2006nsk/Honours%2520Articles/aerodynamics%2520of%2520spin%2520and%2520no%2520sp in%2520tennis%2520bals.pdf+Aerodynamics+of+spinnin g+and+non-spinning+balls&hl=en&gl=ch&pid=bl&srcid=ADGEESiSV4 fZ4EtKnBx_UUTKDtt2EEY3X0o6fk6_pOaXrHKU-uRkIkTDwOvOBqkeROSDmBpgnFb381bO93ToD5mx1-wDF3H9v6TYw9n4-_2AY3ebfcQLW4FQKXEJ7ub_42E3Moi9xyva&sig=AHIEtbQ0MY W4W_GOpm8TGJrw55LKXkOeDw

[tennis_chiro]

not to scale

Here you go: the blue line... still same result...



Uploaded with ImageShack.us

Phil,
That's a good illustration,but not to scale.

Let's see if I can get this one to show

[img]ScaleServerDrawing4FtIn.jpg[/img]

well, preview says it shows even if it is a little too big.

The point is, as you reach in to serve you would clearly have an advantage if you could reach above the minimum trajectory necessary to get the ball over the net and into the court. Obviously, I am engaging in an oversimplification, but let's just say I had a gun and I could get a closer shot at a target behind a barrier, I would want to do it, but if, as I moved up on my target, that target fell protected behind the barrier because I was no longer high enough up to see over it, I would not try to move up. Not unless I was special forces with a GPS guided smart bullet!

My drawing excludes everything but simple plain geometry. It also reminds us of how much we need gravity and the Magnus effect and even drag to get the serve in the service box.

I gotta go teach. More later.
don

[uspta146749877]

I cannot access a file above

Don,
clicking your last attached drawing is producing a message


uspta146749877, you do not have permission to access this page. This could be due to one of several reasons:

1. Your user account may not have sufficient privileges to access this page. Are you trying to edit someone else's post, access administrative features or some other privileged system?
2. If you are trying to post, the administrator may have disabled your account, or it may be awaiting activation.

[uspta146749877]

For Don

Your quote
---->
well, preview says it shows even if it is a little too big.

Don,
I do NOT know what you are talking about

[gzhpcu]

Don,
I feel we have gotten sidetracked on a purely theoretical point, as I have already said. We have seen the effect of gravity, drag and the magnus effect on the trajectory of the ball (the latter even causes the ball to swerve to the side, lengthening the distance between impact on the racket and hitting the service box).

In respect to your original question on tossing the ball far into the court:
- the distance to the receiver is reduced.
- the receiver has less time to react.
- the speed gained on the serve by the severe body lean (more body weight going into the serve) is minimal
- the impact on the biomechanics of the serve is (to me) unclear at this point
- the major advantage today is for doubles players wanting to get to the net fast

Would you agree?