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I don't know how to calculate a tennis racquet head size...thanks!

by YKD
Tags: head, racquet, sizethanks, tennis
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YKD
#1
Dec6-07, 05:58 AM
P: 7
Hi everybody,
The area of a tennis racquet head size is like an egg shape, not oval shape.
how to calculate it?

Thank you so much for your help ^_^
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HallsofIvy
#2
Dec6-07, 06:11 AM
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Unfortunately there are just too many different shapes for an "egg shape" for there to be any good formula. I think you will have to approximate the area by treating it as if the tennis racquet head were an oval or an "ellipse" for which there is an area formula: [itex]A= \pi ab[/itex] where a and b are half the distance across the racquet head measured where it is narrowest and widest.

In other words, you are really using the formula for area of a circle, but, because an "ellipse" does not have a single "radius", instead of r2, you multiply together the values for smallest "radius" and largest "radius".
YKD
#3
Dec6-07, 06:28 AM
P: 7
Thank you, HallsofIvy ^_^
The headsize of my racquet is 89.5 square inch that someone said
but the brand said 93.
Using W x L x 0.8 is closer to this area.
I should use 22/7 x a x b or W x L x 0.8 to caculate the racquet headsize?
http://www.tennis-warehouse.com/view...html?PCODE=LMP

I'd like to calculate it by myself.

Thanks!

Invictious
#4
Dec6-07, 06:34 AM
P: 61
I don't know how to calculate a tennis racquet head size...thanks!

Quote Quote by YKD View Post
Thank you, HallsofIvy ^_^
The headsize of my racquet is 89.5 square inch that someone said
but the brand said 93.
Using W x L x 0.8 is closer to this area.
I should use 22/7 x a x b or W x L x 0.8 to caculate the racquet headsize?
http://www.tennis-warehouse.com/view...html?PCODE=LMP

I'd like to calculate it by myself.

Thanks!
Yes that would be a good idea.

Another idea to do might to be use Integration (just plot half the racquet onto the aces, find the area under the curve, then multiply by 2), though I'm not sure if you have done that before, but either way, it probably will be more accurate.
YKD
#5
Dec6-07, 06:47 AM
P: 7
Quote Quote by Invictious View Post
Yes that would be a good idea.

Another idea to do might to be use Integration (just plot half the racquet onto the aces, find the area under the curve, then multiply by 2), though I'm not sure if you have done that before, but either way, it probably will be more accurate.
It seems quite complicated...
Almost forget to add:
My racquet stringbed (hoop not included)
length:12.25 inch, width 9 inch

Could you please help me calculate it from the link I posted?

Thank you ^_^
CRGreathouse
#6
Dec6-07, 08:25 AM
Sci Advisor
HW Helper
P: 3,684
Quote Quote by YKD View Post
My racquet stringbed (hoop not included)
length:12.25 inch, width 9 inch
I took the picture and put an ellipse with that ratio inside the stringbed and found that it fit fairly well. If it touches on the top, bottom, and two sides there's a tiny bit of space on the diagonals (thus egg-shaped rather than elliptical) so whatever area we calculate should probably be rounded up sooner than down.

12.25 * 9 * pi / 4 is about 86.6, so I'd say an area of 87 or 88 square inches wouldn't be out of line.
Attached Thumbnails
racquet.jpg  
YKD
#7
Dec6-07, 06:52 PM
P: 7
Quote Quote by CRGreathouse View Post
I took the picture and put an ellipse with that ratio inside the stringbed and found that it fit fairly well. If it touches on the top, bottom, and two sides there's a tiny bit of space on the diagonals (thus egg-shaped rather than elliptical) so whatever area we calculate should probably be rounded up sooner than down.

12.25 * 9 * pi / 4 is about 86.6, so I'd say an area of 87 or 88 square inches wouldn't be out of line.
Thank you very much, CRGreathouse ^_^

Perhaps this racquet is difficult to calculate,right?
http://www.tennis-warehouse.com/view...l?PCODE=RQIS1T


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