View Full Version : Tension differences relative to gauge differences

10-16-2006, 07:35 PM
I was wondering if this is a reasonable theory - a racquet strung at 60 lbs with a 1.35 mm string would have the same stringbed tension (and identical pitch when the strings are struck) as the identical brand/type of string in the same racquet but in a 1.30 mm gauge strung at 57.8 lbs, 1.25 mm gauge at 55.6 lbs and 1.20 mm gauge strung at 53.3lbs.

The tension differences are calculated purely on a percentage basis relative to the difference in the string gauges given that the thinner the string the higher the pitch it gives when strung at a given tension when compared to a thicker string strung at the same tension.


10-16-2006, 08:00 PM
It's actually more complicated than that. It's based on a Mass/unit length equation:

TENS = tension
FREQ = frequency
LENG = length
MPL = mass per unit length,
and sqrt = square root

the fundamental relationship is:


where TENS is in kilograms (kg), FREQ is in cycles per second (Hz), LENG is in meters (m), and MPL is in grams per meter (g/m).

The hard part is calculating the MPL.


However, since you're holding DENS(ity) constant between the 2 strings, you can eventually come up with a formula that shows relationship between 2 frequencies (FREQ1 and FREQ2) of 2 diameter (DIAM1 and DIAM2) strings. Both LENG and TENS are held constant, so the only variables would be FREQ and DIAM!

10-17-2006, 02:51 AM
The thinner gauges have more elasticity, though--so I'd go up the thinner the gauge, and do.

SW Stringer
10-17-2006, 08:24 AM
In the book "The Physics and Technology of Tennis", available at TW, the authors make the claim that the stringbed "feel" is directly related to the stress on the string. The term stress is an engineering term and is equal to the tension divided by the cross sectional area. Units are pounds per square inch. To make the stress equal on a given frame with different gauges of the same string use the ratio of the string diameters squared. For example: (1.25/1.3)**2 = 0.925, and (1.20/1.25)**2 = 0.922 . So a 1.30mm stringbed at 57.8 pounds ref tension would have the same stress as a 1.25mm stringbed at 57.8*0.925 = 53.5 ref tension and would be the same stress as a 1.20mm stringbed at 57.8*0.925*0.922 = 49.3 lbs ref tension. The stress in each of the above examples is 28,094 lbs/sq in., and YES that's over 28 thousand pounds per square inch.

10-17-2006, 05:29 PM
In days of yore (1980's), USRSA said to drop tension a bit when going to a thinner gauge, such as 2# going from 16 to 17. I recall the concept being analagous to resistance in electrical wire ... thinner/less resistance/drop tension. Later, the word changed to keep the tension the same.