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12252010, 01:54 AM  #41 
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Really great post .
Extremely useful to all dropweight stringers that are being told off from others for their machines not being reliable and accurate. I use a Pro's Pro challenger I machine with 6 points mounting and fixed clamps and i get the same DT results over and over again while friend of mine that strings on local stringer (older electronic one) gets different DTresults (+/ 1) !!! Many thanks. 
12252010, 07:57 AM  #42 
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Very good post and explanation. As an aside, when using dropweights, you do want to release the dropweight by hand above the horizontal. Let the dropweight stretch the string (a lot or a little if poly/kevlar) to arrive at its desired horizontal position. Then clamp. Clamping too early is not good. You end up with inconsistent results.

12252010, 02:23 PM  #43 
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Best post regarding drop weight machines.
Very good! Thank's 
12252010, 05:52 PM  #44 
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I just gave my ATS/Gamma drop weight machine a hug! I'm also giving her a gift a starting clamp! Happy Holidays!!
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12262010, 03:29 PM  #45 
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Just finished stringing my 18x20 PSC using the 1 piece stringing technique and the result was awesome! This is just my second time stringing using my ATS/Gamma stringer. 'Got much better keeping my drop weight horizontal developing my own technique of using the rachet. I like the over all finish of the 1 piece stringing with of course 2 less knots but it was a hassle handling the longer strings on one side but it gets better as I got closer to the finish line. I also stayed true with my 56 lb tension and it felt right on. Now if only the rain will go away...
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12262010, 06:43 PM  #46  
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Agreed! OP's thread is excellent. Been awhile since he's posted in this subforum, however
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12262010, 07:10 PM  #47 
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Thanks Bud! I should have listened to you sooner though, all that wasted money!
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12272010, 06:39 AM  #48 
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I was just using my brand new dropweight machine and admiring the design. Due to the design, there is very little tension variance between the bar being horizontal and a little off horizontal. As the OP pointed out, the tension is a function of the cosine of the angle off horizontal. But it's useful to note the small angle approximation for cosine is 1((x/57.3)^2)/2. (where x is measured in degrees). Because the angle off horizontal is divided by 57 and then squared, for small angles the cosine is relatively invariant to the angle (because you're on the flat segment of the sine curve).
Approximate tension deviation (in percent) for error off horizontal (whether up or down): 1 degree: 0.015% 5 degrees: 0.38% 10 degrees: 1.52% A 5 degree tilt is extremely noticeable because one end of a 2 foot bar would be about 2 inches higher or lower than the other at 5 degrees (and about 4 inches at 10 degrees). The drop weight is a simple and elegant design for achieving a precise tension. 
12272010, 07:18 AM  #49  
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12272010, 07:28 AM  #50  
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12272010, 07:36 AM  #51 
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I'm about the same age and used to live there in the 80's. Played a lot of tennis with Jim Savage from Okemos. I think he still plays either out of the MAC or Court One.
When did you previously play before your extended break? 
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02132011, 05:08 AM  #52 
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i just got a drop weight machine with linear gripper from eagnas, i made a post in this forum but figured i would ask here about my problem.
could the linear gripper act as a counter balance and effect the final tensions. example if the gripper finishes at 9 o'clock could it be enough of a counter balance to need less tension on the string to keep the rod level. i have been getting varying results with my digital scale from 18lbs to 23lbs depending on the final resting spot of the gripper. (weight set at 20) all other parameters the same and scale not being removed from mount between tries. cory 
02172011, 02:06 AM  #53 
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Didn't know how accurate a regular drop weight could be. Given the accuracy, is there any advantage in getting an automatic drop weight over a regular drop weight? Thanks.

02222011, 04:04 AM  #54 
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You mean something like those "Stringway" machines ??
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02222011, 05:24 AM  #55 
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Yes. I'm now comparing a Stringway ML100 with an Alpha Pioneer DC. The difference between the two is more than $300. If the Alpha can come really close to the accuracy of an auto drop weight, I think I'll save some $ and get the Alpha...
Thanks for your help! 
02222011, 08:50 AM  #56 
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Quite interested in the answer to that one as well. The difference is even bigger over here...an ML100 is almost 3 times as expensive as an acceptable fixed clamps/6 point mounting system "normal" dropweight.
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02222011, 09:40 PM  #57 
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The M90 is also automatic and much cheaper!

03262011, 12:25 PM  #58 
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I just have one thing to say to the OP: Go Bears!

01172012, 04:53 PM  #59 
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Quick question for you. I'm taking ap physics b (ninth grade) and I learned torque to be F x d x sin fee. Why did you use cos?
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01172012, 06:27 PM  #60 
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You pick Cos or Sin depending on what angle you know (or what angle you want to use). In the original post we use cos because the angle we measure is the angle Q, which is the angle between the horizontal direction and the arm. Cos gives the correct properties for this setup. It has a max value (1) at zero degrees (when the bar is horizontal), and a minimum value (0) at 90 degrees (when the bar is vertical). In comparison sin has a max value at 90 degrees. So if we used Sin in the equation we'd be saying you get max tension when the weight arm is pointing straight up or straight down! We know this isn't true, because if you're really good at balancing, you could balance the weight arm vertically without ANY tension on the string! You definitely can't hold the arm horizontal without tension on the string though!
However, you could also look at the problem from a different direction. If you defined the angle Z as the angle between the lever arm and the vertical direction (instead of the horizontal direction), your tension would be max when Z=90 degrees. Z=90 degrees is the same position as Q=0 degrees. The original tension equation in the 1st post was: f = (M g / r) (cos Q) R If you use Z instead of Q you'd just get: f = (M g / r) (sin Z) R Note that the physics didn't change at all, you just looked at it differently the 2nd time. So you pick Sin or Cos depending on how you look at the problem. It's good you ask the question here, because if you can make sense of this now it will be really helpful for you in your AP class. A lot of times you'll learn an equation one way (with sin), but you'll be given a problem where the angle you're given means you need to use Cos instead. Try to do a quick "does this make sense" check when looking at these problems. Ask yourself "what happens when the angle is zero?" or "what happens when the angle is 90?" If the value is biggest at zero, you know to use cos. If it's biggest at 90, you know to use sin. So in this case, you ask "what happens when Q is zero?" well the bar is horizontal and you get your max tension. Which trig function is maxed at zero? Cos. If you asked "what happens when Q is 90?" you'd have to think about it...if Q is 90 then the bar would be standing up vertically. If you had it perfectly balanced so that it was standing up all by itself, how much tension would you have to apply to keep it vertical? Zero lbs! In other words, your tension is a minimum (zero) when the arm is vertical. What trig function is a minimum at zero? Cos. So you use Cos for this problem when given angle Q. Note, if you are dead set on using sin in your equation, you still can't use Sin(Q), but you can use Sin(90Q), since that's the same thing as Cos (Q). It's just whatever way you're most comfortable with, or what makes the most sense to you.
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