Out, out damned spot.(JMac, about 1970) [Archive] - MensTennisForums.com

Out, out damned spot.(JMac, about 1970)

tmurphy
10-02-2006, 12:58 AM
The"shotspot/sureshot/hawkeye/ is a) quite imperfect, and b) grossly abused.
There is not a lot of public information, but we are shown a circular or egg-shaped "shadow" of the computed image of where the ball hit the ground in relation to the ball. But the rules of tennis say that a ball must, at worst, TOUCH the line to be in.
Try squeezing a tennis ball as hard as you can. You can just make a dent which is about 3/4" in diameter and no more than 1/2" deep. Thus the ball only TOUCHES the ground over a 3/4" diameter circle, rather than over the area shown by the shadow. (Who knows if the ball was in? The Shadow knows??) For the spot to be correct as it is used now the ball would have to be completely flat, a physical impossibility, since a) the ball is too strong, and b) where would the air in the ball go?
Time and again I hear the commentators say yup, it just touched the line. Unfortunately if the shadow just touched the line, then obviously the ball did not, because it cannot become perfectly flat, even if Roger Federer is hitting it. In fact at a crucial point, I think a Roddick/Federer tiebreak, Federer's shot was out for this reason, but was called in.
Please don't tell me that the pros hit so hard that it could flatten, because the pros hit the ball very hard, but parallel to the ground. The force which flattens the ball is the vertical velocity, mostly from the simple effect of gravity, plus the top-spin.
I'm not saying the technology is useless, it can determine if the ball TOUCHED the line in certain cases, e.g. if at least half the image of the ball overlaps the image of the line, then some part of the ball touched some part of the line. If no part of the ball image overlaps the line image then the ball is clearly "In" or "Out". Other than that we do not have enough data to make the call. It would be possible to determine the degree of flattening of the ball for various velocities and then show the appropriate area of contact using the calculated velocity.
Finally, why do [I]I have to post this? I enjoy the great commentaries from Cliff, Mary, and the Macs, but come on John, you didn't even swing at this one. TWM

ubky
12-11-2007, 06:26 AM
I think the oval shape is more to do with the ball sliding rather than flattening.

gaz
01-10-2008, 10:12 PM
the ball changes chape in the air and when the ball bounces on the floor it flattens into the court like youo said

the programme they use for hawk eye is so advanced (and would have to be for the ATP and WTA to use it) it takes into consideration the speed of the ball the tyoe of ball and the surface and is actually more accurate when it is close rather than when it is more than 2 inches away from the line

gaz
01-10-2008, 10:19 PM
the ball changes chape in the air and when the ball bounces on the floor it flattens into the court like youo said

the programme they use for hawk eye is so advanced (and would have to be for the ATP and WTA to use it) it takes into consideration the speed of the ball the tyoe of ball and the surface and is actually more accurate when it is close rather than when it is more than 2 inches away from the line

fleabitten
01-13-2008, 05:13 AM
Interesting. I hadn't heard this arguement before.
Teh ball does slide and that explains the length of the ellipse. But the elliptical mark is also about the width of a tennis ball when it is round. So, there may be something to your comment. If you stand on top of a ball on a hard court, it seems like the "mark" on hawkeye is wider than the actual touch of the ball to the court.

flea