The physics of a dropweight tensioner
There always seems to be questions regarding how accurate a dropweight tensioner is. In simple words, very accurate.
I will present here the physics governing a dropweight tensioner. Here is a schematic of a string being tensioned by a dropweight machine: http://ltfang.bol.ucla.edu/stringing/stringing001.jpg We know from physics that, in order to rotate an object such as a rotational tensioner indicated in the schematic as a red wheel, you need to apply a torque. torque = force x distance Newton has figured out that force = Mass x acceleration. On earth, each object with a mass will exert a downward force equal to the product of its mass (M) and the earth's gravitational acceleration denoted as "g." Therefore, the gravitational force (Fg) provided by a dropweight is M x g, as indicated with green letters on the schematic. If we also know the distance (R) between the center of the Mass and the axis of the rotation for the tensioner, we can figure out the torque. Remember that torque = force x distance, therefore the torque exerted by the dropweight (on the right side of the schematic) is: torque right side = M x g x R This is why as you move the dropweight farther from the center, you pull a greater tension on the string. When the string is being tensioned, the rotational tensioner does not move. That's because the torque exerted by the dropweight is balanced by the torque exerted on it by the string. Again, the equation for torque does not change: torque = force x distance. On the left side, the distance in question is the distance between the center of the rotational axis and where the string applies its force. In other words, the radius of the tension head is the distance (r) of interest. The force in this case is the "reference tension" imparted on the string. When the tension head is at rest, it means the torque on the left side equals the torque on the right side, therefore we can set up an equation: torque left side = torque right side, therefore f x r = M x g x R Play with that equation a little bit, you get: f = (M g / r) R Mass (M), gravitational acceleration (g), and the tensioner's radius (r) are all constant, i.e. they never change. Therefore, the reference tension is directly proportional to the distance (R). That's why a dropweight tensioner is simple and requires no calibration, and is impeccably accurate. Now you must be asking this question: what if the tension bar isn't exactly horizontal? Here is another schematic to show you the error when the tension bar is not horizontal: http://ltfang.bol.ucla.edu/stringing/stringing002.jpg Remember again: torque = force x distance Gravity "does not know" if the bar is horizontal, but gravity does know the true distance between the center axis of the tensioner and the center of Mass of the dropweight. In other words, the torque for the right side should be replaced by this equation instead: f x r = M x g x L, where L is the horizontal distance between the center of the tensioner to the dropweight. We know from trigonometry how to related R to L, that if the tension bar is resting at an angle Q, then L = (cos Q) R. Thus, to improve upon the original equation, the tension you're imparting on the string (f) at any angle is: f = (M g / r) L , or f = (M g / r) (cos Q) R At Q = 0, cos Q = 1. Therefore, the number you see on the tension bar, if the bar is resting at exactly horizontal, is exactly the tension you're imparting on the string. Cos 5 = cos (5) = 0.996, i.e. if you set the mark at 60 lb, but the bar is either 5 degrees above or below horizontal, the actual tension is 59.8 lb. Cos 10 = cos (10) = 0.985, i.e. if you want 60 lb., you get 59 lb. That's why the advice is that, if the bar is close enough to horizontal, your tension really is close enough to what you want. Two additional questions you guys might be wondering: 1) I use the center of mass of the dropweight to calculate the distance, but the mass of a dropweight isn't at a point. It spans a distance of 3 inches. Does that introduce an error in my calculation? The answer is no. This simplification does not introduce error. If you don't want to take my words for it, I will prove it mathematically. First let's clarify one thing. In everyday life, we use mass terms as force, i.e. 1 pound of force, etc., the reason is that when we say "1 pound of force," it is implicit that we mean 1 pound of mass multiplied by the gravitational acceleration. For instance, 1 kg of force really means 1 kg x 9.8 m/sec^2 = 10 Newtons. Let's simply the equations and use M as force. By definition, torque in integration form is: http://ltfang.bol.ucla.edu/stringing/eq1.gif Also, by definition, the point of Center of Mass (rCoM) is: http://ltfang.bol.ucla.edu/stringing/eq2.gif If you multiply the radius of Center of Mass by Total Mass, it's exactly torque. 2) The tension bar itself weights a few ounces, does that introduce an error? In fact, it does, but the mass of the tension bar << mass of the dropweight, thus the effect is negligible. Also keep in mind that the mass of the tension bar does not change, either. Thus, it is not a variable for your stringing machine. The error can simply be corrected in my equation by adding a small correction to the Mass, by taking into account the mass of not only the dropweight, but also the tension bar. 
Thank you for posting this. So many times ive wanted to say something along those lines but couldn't be bothered to write it out...
Seb 
Another thing people ask is, what if I push down the bar to the horizontal position?
I took an quick measurement with my dropweight machine. The radius of the tensioner is about 1.2 inches, and the length of the tension bar is about 19 inches. If I apply a force of 1 pound at the end of the bar to keep the bar horizontal, I'm overtensioning the string by nearly 15 pounds!!! How did I come to this conclusion? Look back to this equation: f x r = M x g x R If I apply 1 pound at the end of the 19inch bar, I will add an extra torque of 19 poundinch on the right side (torque). That's the same amount of torque the left side has to come up with. Since the "r" for the left side is only 1.2 inch, the extra "f" it will have to have is about 15 pounds (the ratio of the 2 radii: 19 inch divided by 1.2 inch)! If you have a stiff poly and you're pushing down by a force of 2 pounds, you're overtensioning by over 30 pounds! It can easily break! 
Awesome! This should be a sticky...
Cheers, Tim 
bravo! Well done!

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There is no “error” introduced into any of this by the bar. The total clockwise (in your drawing) torque applied to the drum is equal to the sum of two torques, the weight of the bar times the horizontal distance of the C/G of the bar from the center of the drum, plus the weight of the weight times the horizontal distance of the C/G of the weight from the center of the drum. The two torques go up and down together by the same percentage, with any change in angle of the bar, since both C/Gs move back and forth horizontally together by the same percentage, so there’s no error being introduced. 
^^^
You're right. The tension bar applies a constant torque which is always there. It simply adds to the torque from the dropweight (which is by far greater). I just meant that, I neglected that part when I wrote down the equation using the dropweight as the sole source of pulling force. Indeed, it does not introduce an error in the actual stringing of the racquet. 
lovely write upI'm wondering, though, if ever to let the weight "freefall" even for only an inch or two. The correct method is to keep the weight in hand and move it back to horizontal without it ever leaving the hand? Since you say that putting a 1lb force on the top of the weight can translate to a 30 lb difference, I'm a bit curious because sometimes I'll press downward on the lever at the last string to compensate for slack, a bad habit I realize after reading this

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Allowing the dropweight to drop a little bit shouldn't be too much of a problem. The elasticity of the string will pull back the dropweight, allowing everything to go back to the resting position. However, to drop it from a high position, the dropweight will pick up velocity as it drops down unimpeded at first, and it may create too great of a force and it can actually break the string. Remember, Power = Force x Distance / Time. If you allow the dropweight to drop too fast, it will create great power. It may be possible to break the string if you allow the dropweight to free fall from its highest position. No wonder big hitters break strings all the time. Edited to add a better explanation: if you drop the dropweight from its highest point, it will pick up speed as it drops down. In other words, the dropweight is picking up kinetic energy. When the dropweight finally reaches its lowest point, the kinetic energy it has picked up will have to go somewhere: the string and the frame. If the strings cannot handle that energy, it will snap, and let's just hope your frame is still intact. That's why you need to allow the tension bar to drop slowly. 
Excellent post, easy enough for me to understand even though I'm quite mathematically challenged :). Sticky anyone?

Great post. Theres a lot dropweight users out there. I agree, this should be a Sticky......

If you are not already one, you will make a great engineer :)

definitely knew about the velocity part, but just to clarifyyou suggest keeping the weight in hand at all times, right? This minimizing the effect of velocity affecting the accuracy of the job?

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For each pound you push down, you overtension by about 15 pounds. Likewise, for each pound you push up, you undertension by about 15 pounds. 
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the effect of the tension rod weight is not negligible. when i got my dropweight, being the nerdy engineer i am, i had to verify that the tension scale on the rod was correct. i took all the above measurements and to come up with my own solution to where the weight should be for a desired tension. i KEPT coming up answers different than the tension scale on the machine. i kept thinking about what horrible engineers must have designed my machine. THEN i realize i forgot about the moment produced by the weight of the tension rod. abracadabra all the numbers mached! and thus, the tension scale on my machine WAS correct! it turns out the tension rod (on my machine) contributed about 56 lbs of tension! ... definitely, not to be neglected. this is a bias, so no matter where the weight is along the rod, the tension rod itself always adds about 56 lbs. in addition, you need to be very careful when tweaking the mass of your dropweight to account for other error sources. your should NOT do this to account for the tension rod weight. this is b/c the tension rod mass adds a constant bias to the pulled tension, but the mass of the rod affects the tension via a scalefactor (i.e. the distance R). the bias is a result of the fact that the tension rod mass is fixed, where as the scale factor arises b/c the dropweight mass moves. different error sources, don't mix 'em. 
If you use a dropweight should you get the push down bar to balance correctly or should you hold it up to the correct tension with 1 hand and clamp with the other

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Good that you've pointed it out.
I've never really considered it in detail, since the tension bar itself applies a constant torque, which means the scale bar is moved up uniformly, a correction for the manufacturers to make. Quote:

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otherwise, i feel very bad for you, your friends, and anyone else for whom you string racquets ... 
lol i do let gravity do its job. I also have only strung 2 rackets for self.

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