Physics of V-Stretches

I understand the physics behind why V-stretches are supposed to be better, but it seems flawed to me. I’ll do this unnecessary detail, just for fun.

Take a cable, suspended between two telephone polls. Pull down on the cable half-way between the telephone polls. We have a situation similar to a V-stretch. Assume ideal conditions (zero gravity, two dimensional cable, etc.) Now, take an incremental length of cable from any point, and it is necessarily acted upon by two tensions of equal and opposite magnitude acting along its length (otherwise this section of cable would be moving). Move along the cable one incremental length at a time, and you’ll find the tension in each section of the cable must be the same as every other section.

We don’t need to analyze the fulcrum point, but, summing vectors: Cable tension acting towards left hand telephone poll joint + Cable tension acting towards right hand telephone poll joint + force cable fulcrum is being pulled down with = 0.

So the tension in every single point of the cable is equal. This is pretty basic mechanics.

Now for my point of contention. What is the magnitude of the tension in the cable? Take the incremental section of cable with one side joined to the left telephone poll. It is being acted on by two forces, one, the tension in the cable, two, the supporting force from the telephone poll. The section is not moving, so, talking in absolute values, the magnitude of the tension in the cable is equal to the magnitude of the support force at the telephone poll. Let me repeat that. The magnitude of the tension in the cable is equal to the magnitude of the support force at the telephone poll.

In other words, the tension in the penis is equal to the force with which you pull it! Exactly the same as if you were doing a normal manual stretch!

Now, the changed shape and bend of the penis may cause the stretch to emphasize different areas in different ways, but do not be confused into thinking the force in the penis is magnified by adding a fulcrum. Assuming a stationary penis, the tension it supports is always equal to the force with which you pull it.

Yes?

You are able to achieve greater force by adding additional tension from the fulcrum point. Try it, you’ll see. The analogy is flawed; the support poles are merely holding the weight of the cables, if they were pulling the wires downward it would increase the tension.

Paging MagnumXL

memento - There is nothing in that thread I do not understand. Attaching the penis to a wall would be a great idea, because the force you push down with would result in a magnified tension in the penis. But if your hand is there instead of the wall, the tension will always be equal to the force that hand pulls with.

soflsul - You are wrong. There are three forces acting on the cable. The two supports and the downwards force in the middle. The two supports are pulling upwards and to each side, the downwards force is pushing down. Just like the base and your hand are pulling back and forwards respectively, and upwards, and the hand-fulcrum is pulling down.

Oh, sorry memento, I read that thread more carefully, it seems you realised what I did and agree with me :) .

As solfsun says, you are only taking into account the simple load. Clearly the forces add to zero because the penis is not accelerating. Take an inverted V-stretch over a baseball bat over your legs, the bat is pushing up with a certain force and you are pulling down with a certain force, the tension is going to be related to the vectored force of both.

Only taking into account the load caused by the hand is a bit like asking why a suspension bridge collapsed only taking into account the mass of the bridge. In fact the maths for suspension bridges are probably a good thing to read up on when studying vectored force.

EDIT: I kind of agree. I think that using the example of two fixed points and a force bearing on the centre is erroneous, I think its more complex than that.

Well, you’re both still wrong.

The mass of the penis is much less than the forces exerted, so it can be ignored and we can still get a pretty good approximation of what’s going on.

Here’s a jpeg I just did, maybe it can explain better.

Adding to the jpeg,

H = F/2sinx

Meaning, in order to support a certain fulcrum force, you need to pull with your hand with a force equal to the fulcrum force divided by twice the sin of x.

By your explanation, F only has a Y component and is therefore 2Psinx (<—the angle), So the force causing tension is not P but 4Psinx(<—the angle). X components cancel.

EDIT:
Damn you edited while I was writing.
>Meaning, in order to support a certain fulcrum force, you need to pull with your hand with a force equal to the fulcrum force divided by twice the sin of x.<
Effectively we are saying the same thing. Its not P/H is the point I guess.

If nothing else, the fulcrum will give you more leverage and increase the force with which you are able to pull. If the fulcrum enables you to pull harder, than it has increased the tension. The point still remains, V-stretches allow you to increase tension. Keep reading Nedd.

… What?

F = 2Psinx. Yes.

So the force causing tension is 4Psinx? Twice the fulcrum force? 4 times the y component of P? Why? Read my first post again. The tension in the penis has to equal the support force, i.e. H and P.

I don’t think we are saying the same thing.

H = F/2sinx. But H still equals H… The tension in the penis still equals H…

How does the fulcrum give you more leverage?

Okay Nedd, let me explain it a little more clearly; when I add a fulcrum to my SO stretch, it feels better. I feel a greater tension. If this is imaginary, I don’t mind. Thank you for the formulas.

>So the force causing tension is 4Psinx?
Sorry I’m wrong here, the force is

4Psinx + 2Pcosx

Just because X components cancel they can’t be ignored, silly me.

The forces involved are F,H,P. F only has a Y component, as Y components of P & H are together equal and opposite to F The full force is 2F or 4Psinx.

The X components are also equal and opposite, hence the 2Pcosx.

This is confusing, X components and x the angle. Couldn’t the angle be a or something if its not theta.

Okay. Just trying to clear up the confusion. I’d suggest it feels better because you’re pulling harder.