Where is the tension applied in a 3 dimensional cable?

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I have a pulley with a radius of 5 mm and a rope with a diameter of 0,75 mm. I've judge that the maximal radius (pulley+cable) is too different of the radius of the pulley and that I cannot neglect this difference.
Hello,

I need to transmit motion between a servo motor and a pulley through a cable. In order to do that, I've screwed a pulley to the servo and fix the cable on both the motor pulley and the pulley. I am working in a scale where the radius of the cable cannot be neglected due to a precision requirement. I actually don't know with which information I am supposed to used in order to calculate the rotation ratio of this system. I'm was thinking that the maximal radius, the combination of the pulley and the cable diameter, was the emplacement where most of the tension seems to be applied on the rope, thus I've used this value on both pulley to find the ratio. Some experimentation proven me wrong. Does someone know how to solve this kind of problems?
 
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I think you need to show a diagram of just what you mean. Also, your pulley is really just .4" in diameter?
 

tech99

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We really need a pulley to be several times the diameter of the cable to work properly. In terms of gear ratio, I think you must assume that the cable diameter increases the effective pulley diameter. Maybe measure to the centre of the cable diameter.
I have recently been making a school project with a pupil where we need a high gear ratio for a model car, so we have used two steps of reduction so we can use larger pulleys.
 
Am I correct in thinking of your system as a servo with an arm, and on the end of the arm is a pulley wheel, with a cable running through it, affixed at one end to a fixed point and the other end is attached to what you wish to move - the item which provides the resistance, for which you want to measure the tension in the pulley?

Or, is the pulley turned by the servo, and the cable is pulled by the actual turning of the pulley?

Personally I would make all calculations by treating the cable as a zero-thickness line, in the centre of the cable. so take whatever it's resting on and add half the diameter of the cable to the radius. where the cable bends, the inside is compressed and the outside is in tension, so the average on the centre remains the same.

IE, if you want the radius of the cables position, you would add the radius of the wheel to the radius of the cable, this is the distance of the centre of the cable to the centre of the pulley. IE, 5+(0.75/2) = 5.375mm.

If you can provide a quick sketch of how the system is laid out, ideally with measurements, then I'm sure you will get some answers. I'm struggling to work out what I'm supposed to be answering!
 
Here is the schematic of the system. It is the first time that I am using inventor to draw, thus I must apologize for the quality of this one. This draw have dimension that are 10 time the original one and these are millimeter.

243725

For your question:
-Yes, the bigger pulley is just 0.4".
-The pulley is effectively turned by the servo. It is attach to its horn. I also want to find where is the tension applied in order to use this application radius for my rotation ratio calculation.

Another detail, the other end of the bigger pulley is attach to an elastic, which is acting as a tension spring.

I will try measuring from the center of the cable and test this once I'm done printing a new test bench.
 
After some experimentation, it seems that using the sommation of both the cable and pulley radius is effectively an excellent estimation, Since I only obtain value that are very close, I think that it is confirm, even if I did not take a lot of measure.
 

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