Lead Tape/Swingweight Question?

langdon0555

Semi-Pro
Does anybody know a calculation for how much lead tape will increase swing weight by 1 point at a time, if you were to put the lead tape at 12 o'clock on the racquet?

For example .25 grams of lead tape would increase swing weight by 5 points, etc....

Thanks
 
C

Chadillac

Guest
Does anybody know a calculation for how much lead tape will increase swing weight by 1 point at a time, if you were to put the lead tape at 12 o'clock on the racquet?

For example .25 grams of lead tape would increase swing weight by 5 points, etc....

Thanks

I apologize for forgetting the user here that gave me this info

Use the Weighted Mean formula. Assuming that:
- W1 is the weight of your racquet (w/ syn grip) and BP1 its balance point (from the buttcap);
- W2 is the difference in weight b/w the leather and syn grips and BP2 the position of the grip's center-of-mass (always from the buttcap, so it will be something like 7 to 9 cm? just measure how far from the buttcap is the middle of the grip);
- W3 is the weight of the lead tape to be added and BP3 its center-of-mass (same concept as for the grip above)

You have to decide where to put the lead tape. General recommendations are 9 & 3 for increased stability and power, from 11 to 1 for increased power and to move the sweetspot higher in the stringbed, etc.
Now, reversing the Weighted Mean formula (see below), you can find out W3 (how much lead tape weight you need).

Formula:
Final BP of the racquet (you decide it) = (W1*BP1 + W2*BP2 + W3*BP3) / (W1 + W2 + W3)
Solve it to find W3!


EDIT: I had some time so I did it for you

W3 = [BP*(W1+W2)-W1*BP1-W2*BP2] / (BP3-BP)
Put the data in an excel sheet and you're done.

NOTE: this method is for calculating the "absolute" balance point of the racquet and does not take into account the actual pivot point, which depends on your hand position on the grip!

Looked like too much work for me so i felt it out, great info though
 

zalive

Hall of Fame
Does anybody know a calculation for how much lead tape will increase swing weight by 1 point at a time, if you were to put the lead tape at 12 o'clock on the racquet?

For example .25 grams of lead tape would increase swing weight by 5 points, etc....

Thanks

EDIT:
6'' of 1/4'' wide lead tape side put at 12 o'clock (on both sides) will increase SW by approx. 10 units.
Therefore, you need 0.6'' (1.5 cm) * 2 of lead tape at 12 o'clock to increase SW by 1 unit - 1.5 cm length per side, 3 cm of lead tape total.
 
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Irvin

Talk Tennis Guru
Does anybody know a calculation for how much lead tape will increase swing weight by 1 point at a time, if you were to put the lead tape at 12 o'clock on the racquet?

For example .25 grams of lead tape would increase swing weight by 5 points, etc....

Thanks
The SW increase will be the distance from the SW axis to where you put the lead tape squared times the mass in Kg. Most people use a SW axis of 10 cm that's pretty standard. If you have a 27" racket and place the lead inside the hoop at 12 o'clock your distance is ~58 cm. so 1 (SW) = 58*58*kg Do the math for a 1 point increase it takes 0.3 g SW = 58*58*0.0003

EDIT: Another way to say it is the the reciprocal of the distance squared from axis to lead will increase the SW 1 point.
 

bageldog

Semi-Pro
Does anybody know a calculation for how much lead tape will increase swing weight by 1 point at a time, if you were to put the lead tape at 12 o'clock on the racquet?

For example .25 grams of lead tape would increase swing weight by 5 points, etc....

Thanks
sorry everyone is making this so difficult for you. 1 gram at 12 is around 3.3 SW units, and 1 gram at the 3/9 position is about 1.8 SW units. or very roughly...1 gram at 12 is 3 SW, and 1 gram at 3/9 is 2 SW
 

Irvin

Talk Tennis Guru
@bageldog is right everyone is complicating the calculation. Always weight your lead and don't assume it's weight.

SW = ddM where SW is SW increase, d is distance from lead to SW axis, and M is mass in Kg

SW/dd = M

to convert to grams multiply by 1,000

1000*SW/dd = M (in grams)
 

Irvin

Talk Tennis Guru
Follow up question...

How many swing points does an unstrung racquet go to when stung usually? 20 points, 30 points?
That depends on the string, gauge, head size and racket. A stringbed has a fixed size and pattern (usually) but the inertia of the string bed can change from string to string. Inertia is additive, that means any mass you add to an existing racket creates more inertia based on the distance of the mass from a common axis.
 

Irvin

Talk Tennis Guru
@bageldog is right everyone is complicating the calculation. Always weight your lead and don't assume it's weight.

SW = ddM where SW is SW increase, d is distance from lead to SW axis, and M is mass in Kg

SW/dd = M

to convert to grams multiply by 1,000

1000*SW/dd = M (in grams)
Another point SW is inertia pure and simple. All forms of inertia have an axis whether it be Twist Weight, Inertia at COM (Recoil Weight,) Inertia @ COM (Icm,) or Swing Weight. But SW is the only inertia measurement that can be calculated on different axes. SW are not additive unless your using the same axis. For instance if you SW is calculated on a 10 cm axis you can't convert the SW of a 350 g racket to 0 cm using the formula SW = 10*10*.35 and add that to the 10 cm SW.
 

RanchDressing

Hall of Fame
a little less than 30 for a 16x19 with poly, 98-100 head size
This is true, but just want to add, this totally depends on the gauge and string. In a 16x19 pres pro (95), volkl cyclone 16 gave me +31sw confirmed over and over. The measurement for shaped polys seems to be weird however. So just keep that in mind. I wish TW would publish their database of unstrung to strung SW/Weight conversions. It's mindbogglingly large.
 

Irvin

Talk Tennis Guru
It's not really that mind boggling when you consider the addition of strings moves the balance about 1 cm up on the racket and farther away from the 10 cm SW axis. Also the center of the string bed is ~40-45 cm from the 10 cm axis. Mass (16-19 g) at that distance will raise the SW by 25-35 points.
 
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