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Thank you in advance to anyone with time to take a look at my question here. For anyone interested, this is a question in regards to an open source film scanner project, which can be found here: https://kinograph.cc

I have a machine that pulls a material through rollers (in my case, a reel of movie film), and I want to maintain constant tension. I've added a roller that can measure load and provide it to the system to adjust accordingly. My question is about the placement of the rollers and how it would affect my readings over time.

In this first diagram, I simplify the setup to better illustrate the question, which is: how does the angle between two rollers affect the force measured on a single roller? In the second diagram, I show the actual use case and how it is effected by the overall diameter of the spool as it collects the film at the end, relative to the location of the roller.

My hunch (as a non-engineer) is that the "A" diagrams would provide a consistent reading of force on the sensor roller throughout the duration of the spooling and would thus be a better choice for a feedback loop. In this case, I could just set a target value of stress on that middle roller and adjust the rest of the system to maintain that value (like a PID loop).

Am I correct there? And also that the second version would result in a "moving target" of force on the sensor roller that would be harder to maintain within the system (for an intermediate programmer with basic math skills like myself)?

simplified diagrams actual application diagrams

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To my understanding, your system of rollers and material is equivalent to that of a pulley and a rope. Doing some research here might give you answers for future questions.

Since you have multiple questions, I will first give a summary of my point of view, before going more in-depth on the details of why.

Summary Yes, your hunch is correct. The parameters of drawing A will be more consistent throughout the duration of the spooling, and will hence also make it easier to write a script interpreting the forces affecting the roller and provide it as feedback to the system.

Assumptions Before heading into your example drawings, let's establish some assumptions. Since you are the one with the most insight to your project, I suggest you consider these yourself, determining if they are in fact applicable to you.

  1. There are no friction between the roller and the material passing over it. Why is this important? Because it ensures that when pulling the material from one side of the roller, the exact same force will be applied to the material on the other side of the roller, only in the opposite direction.
  2. The material does not stretch (notably, that is).

How does the angle between two rollers affect the force measured upon a single roller? I interpret this questions as: How does the change in starting angle and ending angle of the material passing over the roller affect the force on the roller? The simplest way of showing that it will have a large effect is by the trivial case of a 180 degree change and a 0 degree change, as shown below: Trivial case of force on roller

However, this is a trivial case and thus not very useful besides displaying the extremes. Let's have a look at a more general example, as displayed in the image below. Let $\alpha$ represent the change in angle of the material passing over the roller. Assuming a constant force $F$ pulling on the material, it is the value of $\alpha$ that decides the resulting force $R$ acting on the roller, as well as the direction of the force.

To compute the resulting force, we first need to decompose the forces acting on the roller as shown in the right image below. This is a quite simple task as long as you make sure to use the correct trigonometric functions. Note how the angle $\alpha$ is found twice in the decomposition, once on the bottom left and once on the top right. This is possible to due the parallel lines (visit this page to learn more about angle similarity for parallel lines).

Decomposition of forces

When the decomposition is performed, it remains to compute the sum of the forces in each direction as:

x-direction: $F = R_x + F * \cos\alpha \implies R_x = F * (1 - \cos\alpha)$.

y-direction: $R_y = F * \sin\alpha$

To compute the resulting force, we then use the Pythagorean theorem, giving $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{[F * (1 - \cos\alpha)]^2 + [F * \sin\alpha]^2} \\ = F * \sqrt{1 - 2 * \cos\alpha + \cos^2\alpha + \sin^2\alpha} = F * \sqrt{2 - 2 * \cos\alpha}$$ where we have used the trigonometric identity $\sin^2x + \cos^2x = 1$. When knowing $R$ you may also calculate the angle of attack using any trionometric function.

Note how I in the above example chose to keep track of the direction of each force using arrows in my drawing. However, another option is to leave out all arrows and stricly use '+' and '-' relative to a defined coordinate system. Both methods are valid, but if performing the computations using code, I would probably suggest the method using signs.

Which drawing is better? As just shown in the example above, the angle $\alpha$ affects both the resulting force and it's direction. If choosing drawing B, you would have to account for the change in alpha at all times. On the other hand, drawing A will always have the same input and output angle due to the rollers next to it, meaning that the direction of the force will be constant. This makes it a lot easier to measure the force and perform the inverse calculation as above to get the force $F$ acting on the material.

(Ref your top right figure) If pulling blue material with same force as the left figure, will it produce less force? Yes, since the angle $\alpha$ is smaller.

Does the exit angle matter (in orange)? Judging by your figure, no. This is because the roller next to it will make sure the input angle to the roller which is measuring force will always be alike. However, the angle becomes so large that the blue material does not touch the first (outmost left) roller, then it will have an effect.

As a final note, I would highly urge you to go through my calculations yourself to verify that it is indeed correct (the thing we all have in common is our ability to make mistakes), and that you understand it sufficiently enough to use it in your project. Good luck!

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  • $\begingroup$ thank you so much for this thorough and thoughtful response! much appreciated. $\endgroup$
    – mepler
    Apr 30, 2022 at 16:18

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