reduce the main string 2 lbs when a customer wants re-string 23lbs, it's mean main 21lbs and cross 23lbs, but not all cross string is 23lbs, it's from head's 6th staring string, it's mean from the sweatspot to bottom.
String tension Hello everybody, I'm stringing my racket also with 5-10% less tension on the crosses. Somewhere I was reading an Article in the net that the recommendation ( stringing the crosses 2 lbs higher) is only because the deformation of the headshape. The main aim should be gaining an equal stringbed tension. Therefore because the crosses are shorter than the mains they have to be strung with a lower tension. I tried to proof this with some physics. Here are my calculations Because the string lenghten itself while under tension the diameter => the cross sectional area of the string is getting smaller. A = V /(L1 * k * F1) ; A = Cross sectional Area, V = Volume of the unstretched String, k = lenghtening of the string (%), F1 Force (N) which pulls on the string the tension of the string is coupled with it's lenght through: 2 * L1 * f = sqrt(F1/ A * ro) ; f = Frequency, ro = density (constant) (c = sqrt(F1/ A * ro) ; , c = propagationspeed of the wave with c= lambda f; lambda = wave length = 2 *L1) The frequency ( Sound) of the long (main L2) and the cross (L1) string should be the same, therefore we can insert the two equations into each other and get : L1/L2 = sqrt[(F1* F1 * L1 * k * v2 * ro)/ (F2 * F2 * L2 * k * v1 * ro)]= sqrt([v2 * L1]/v1 * L2]) * F1/F2 we can cancel some terms and get: => F1 = F2 * (sqrt(v2 * L2/ v1* L1))^-1 ; with v1 = (d/2)^2 * pi * L1 , d diameter of the string the equation simplifies to: F1 = F2 * ( L1/ L2) If we take L1 = 18,8 cm for the cross string and L2 = 23,5 cm for the main strings we get a tension ratio of: F1/F2 = L1/L2 = 0.8 If the mains are strung with 22lbs (10kg) the crosses have to be strung with 17.6 lbs (8kg) to get the same overall stringbedtension. Of course this is just for the longest strings, but I think it should be clear now that the crosses should have a lower tension. Hopefully there are no mistakes in my equations but the result is just to nice to be wrong. Lets hear what you think about it. Regards mark
Why 2 more lbs on the crosses Hello, it seems as nobody is reeding my post.:crying: So I had to find an error myself. The results were just to nice I think I have an error in the equation to calcualte the cross sectional area A in dependence of the Force, because if you apply zero force the length of the string would be zero! A = V /(L1 * k * F1) ; A = Cross sectional Area, V = Volume of the unstretched String, has to be corrected to: A = V /(L1 * (1+ k * F1)) the rest follows analogue to my previous post. But the result is not as nice as before. The equation for the wanted string tension (Force) is: F2 = 1/2K2 * [sqrt(1+4k2(L2R2/L1R1)^2*F1*(1+kF1))-1) Hopefully the equation is now correct If we are using the relative lenghtening from Neils post for BG 65 (0,70cm) 17% and 23.5cm for the mains, 18.5cm for the crosses and the applied force is 100 N ( roughly 10kg = 22lb), the the tension for the crosses should be 78,1N or 7,8kg which is nearly the same ratio as before (0,8) With the computed equation it's also possible to calculate the necesseary tension for the proportinal stringing method or the resulting tension if you are using an other string. For example if you are using normally the BG65 and string this one with 22lb and use this tension for BG68TI the equivalent tension with BG65 would be 31 lb!! Due to the extreme lenghtening of the string 33 %. If there are still some errors I would be glad to get some corrections. Regards mark
the figures I got for lengthening were 1.17% per kg, not 17% so for 10kg tension, the string would stretch 11.7%
why 2 lbs more on the cross Uups, you are right. Ok with the new value I get 7.3 kg on the cross. Perhaps someone is able to verify the results. Regards mark
hi mark all u have done was deriving the basic ratio quotient the long way x1/x2=y1/y2 actually, equal main and cross tension is just one of many objective desired by the stringer/user
i just popped into this forum section after reading the equipment section... and i noticed this thread... adding 2 pounds to the crosses to maintain the stringbed's normal shape so it isnt deformed. therefore, i'm going to string my new mp99 with bg66@main 22lbs and cross 24lbs? but what will the tension be? both equal since the mains are longer? therefore both will be at 22pounds, or what? and also the mathmatical equation by baumbaer and neil. to put it in short,the conclusion you have reached is that if one kind of string is done at a certain tension, another string will have to be at a different tension to be exactly the same tension after stretching? correct? very complicated indeed...and im making no sense of that mathmatical equation...
My experience of playing with and stringing BG66 make me expect it to stretch the least. But I'll try to remember to test some next time I open a packet. (I think i've got some now)
i want my racquet to be strung at 24 lbs even therefore, with this pattern, i would string it at 24@ mains 26@ crosses after a couple ours of rest, then it would relax to 24 together? btw, is the mp99 string pattern 22 mains 22 crosses?
re-tested BG66 under the same conditions as the 65,80,68 and got 1.47% The BG70Pro figure should maybe be discounted also. I don't know how much those 66 and 70 from a long time ago had already been stretched.
Here's a thought for everyone. Usually when I do string jobs for people, I string with a 2lbs difference between mains and crosses. One of my recent clients however accused me of not doing the 2lbs difference because the racket looked too "fat" (horizontally too wide)-- I assured him that I had done the 2lbs difference and that when I gave him the racket it looked fine. When I went to see him at his gym, expecting to see something wrong, he showed me his AT800DE which looked perfectly normal to me, exactly the way I had strung it... the reason why he thought it was fat was because he was comparing his AT to someone else's that was strung "also at the same tension", but that looked extra *long*. I mean, the racket face was actually thinner (noticibly less wide) and longer (noticibly longer from head of the frame to throat). The owner of this "AT800DE Long" (i just made that name up) suggested to me that I was stringing incorrectly, wheras I am sure that it is his stringer who perhaps made an error. I compared my stringing job to an AT800DE without any strings and the shape of my stringjob is almost identical to a stringless AT800DE. So what I'm wondering is this... the person with that AT Long insists that he had his job done by someone in China who strings for the national team. I didn't pry for details, it's not really important. But what I want to know is, has anyone here at BC ever heard of using MORE than 2lbs difference on mains and crosses? And, moreover, is this safe? My own answers to these questions is, no, I have never heard of more than 2lbs difference. As to safety, i'd imagine that any distortion from the stringless form of the frame could be dangerous to the racket. But i'd like to get some other opinions as well. I am thinking from the look of it that this isn't even something like 'a little more' than 2lbs difference, it's got to be even more than that considering that my first sight of the racket was "holy cow, it looks like a rectangle with rounded edges!" I don't beleive the player who uses this special AT has never really had any problems so far....
check the AT-long. dose the head look more square? imho, even at 19mx24c, my cab20ms is identical to another one with 19mx21c. however my old aerotus88 will change shape. i think the ms box frame will not change as much compare to the thinner defense frame. just like to get some opinions from other people. do you think the aerodynamic shape frame is softer?
Well, personally, I find it's softer than some but stiffer than others. The 'softest' racket i've seen so far is in my opinion the Nanospeeds. On another note-- i'm posting to folow up on this. I did a little experiment on one of my AT 800smanaged to acheive an AT800De "LONG" by stringing the crosses, instead of from top to bottom, from bottom to top. I guess that usually when we use the recommended top-down direction for stringing crosses, the frame of the racket gets "stacked" downwards, and the opposite is true too. The thing I'm concerend about with these 'long' ATs is that basically what they are is a regularly strung AT, but the bottom 'fatter' part is more compressed-- I rather think that this means that the lower area of the stringbead will be more tense than the upper area. So my guess is that stringing form bottom up makes the sweetspot move lower? Any thoughts? Also, does anyone here actually string crosses from bottom-up? Is this bad for the racket?
I've always done bottom up on the crosses because I pre-string a 1-piece string then tension it. Never had a problem.
Hey benibeni, Yonex recently changed their stringing patterns. The Nanospeed 9000, and the Armortec 800 Off & Def are strung differently than all the other Yonex racquets and there was some buzz going around saying that they had changed the tension different of 2 lbs. I work in a pro-shop and had heard this rumour a number of times. I hadn't heard anything from Yonex but decided to phone them up anyways. It still stands, with all the Yonex Badminton racquets they should always be strung with 2 pounds higher. That's coming from the mouth of the Yonex representative at the head office. Anywho, there's my 2cents!
Ummm...aren't all Yonex badminton racquets strung as a 2 piece? At least that is the way that Yonex recommends that you do it. Same with their tennis racquets! You string thru to the 9th string, then go from T12 (twelfth throat grommet) to H11 (eleventh head grommet) over then to H10 and down to T10. Then tie off on T8 or T6. (Can't remember what they recommend). Then you start your crosses from bottom up. You don't string them as a one piece do you?
