I'm no scientist so maybe my assumptions are way off but I wanted to know if I am. If you have 22 strings and each main string of the racket head is at, say 25 lbs tension, does that mean there is a total of 550 lbs of force being applied to that racket head? I assume the reason it doesn't cave in under that much force is because it is being cancelled out by (hopefully) an equal amount of force from the other set of strings (cross). Does that mean the difference in tension between the mains and cross strings will be what is felt by the racket head? It's scary to think if you're off by just 10%, your racket head will have 55 lbs of pressure on it.
There are, in the "standard" pattern, 22 mains and 23 crosses, so if you do them all at, say, 25 lbs, that means there's a total of 45 x 25 = 1125 lbs of compressive force. All in something that weighs less than the average apple... I feel bad for using 29/32 now!
double that. remember that each string segment is pulling on both ends, so the total force on the frame is 2250lbs. enough to pick up a small car!
Now we have a better appreciation of how strong carbon graphite is . . . But no carbon bamboo is strong enough for the Panda's dinking sessions.
Or a Ronnie Coleman leg press. Weight for weight our poor rackets take so much more strain than their fatboy tennis cousins, who weigh four times as much but only have to cope with 2-2.5 times the pressure.
My wife drives a Toyota that weighs less than 2250 lbs - imagining it falling on my precious Ti10, ouch....
Nope, it's not pulling with 25 lbs on both ends. It's pulling with 12.5 lbs on each end, making it a total of 25 lbs. If when you pull the string with 25 lbs force, then that force will be divided evenly along the pulled string
it is 25lbs on both ends. just think about a stringing machine, when i pull 25lbs, i am pulling it with a 25lbs weight. in order for that string to be in equilibrium, the needs to be an equivalent of a 25lbs weight on the other end of that string. so there is 25lbs on each end.
