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

by YKD
 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 ^_^
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 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: $A= \pi ab$ 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".
 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!
P: 61
I don't know how to calculate a tennis racquet head size...thanks!

 Quote by YKD 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.
P: 7
 Quote by Invictious 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...
My racquet stringbed (hoop not included)
length:12.25 inch, width 9 inch

Thank you ^_^
HW Helper
P: 3,684
 Quote by YKD 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

P: 7
 Quote by CRGreathouse 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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