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Physics of V-Stretches

Isn’t v-stretching all about targeting your septum or tunica locally, the spot where you’re applying the fulcrum?

Piet,

Its more about increasing the force.

The tension will be the same throughout the penis.


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>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, this is getting confusing. I’ve done a couple of years worth of university maths, physics, and engineering mechanics papers, with straight A+s. Can we trust that my analysis of an simple free body force diagram is correct?

Ned,

Then you must have caught some newton. If H was the only force acting you’d have acceleration.

Reduce the angle x to zero (ie no fulcrum) P=H therefore the force causing the tension is 2H.
Increase the angle to 90 degrees and F=2P and the force causing the tension is 4P or 2F.


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memento, I think, in addition to making an error in summing the y forces, you’re trying to sum up the x and y components of both forces. You’re not supposed to do that.

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Originally posted by memento
Ned,

Then you must have caught some newton. If H was the only force acting you'd have acceleration.

? H isn’t the only force acting. There are F and P as well.

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Reduce the angle x to zero (ie no fulcrum) P=H therefore the force causing the tension is 2H.
Increase the angle to 90 degrees and F=2P and the force causing the tension is 4P or 2F.

If x is zero, F must be zero else P and H are infinite. The tension in the cable is still equal to P which is equal to H. Attach a cable to a wall and pull with 10N force. The tension in the cable is 10N, even though there are two 10N forces acting on the cable (your pulling and the wall’s pulling).

If x is 90, P = H = F/2 (sinx = sin90 = 1); there are no x components. P is always equal to H though, and these two are always equal to the tension in the cable.

Read my very first post again. The tension in the cable must be equal to the support forces.

Did I make an error? Quite possible of course, where?

So simple stretch:

You pull on your penis with a 10 newton force, because there is no acceleration we have an equal and opposite component from the base pulling in the opposite direction with a 10 newton force. Is the tension 0, 10 or 20 newtons?

With the 90 degree example you are pulling on your penis with a 10 newton force, because the Fulcrum is exerting an opposite effect it must be opposing this force at least equal to H. Its a reasonable assumption that P is also 10 newtons because the penis is not slipping over the fulcrum point. Therefore F is 20 newtons. So is the tension 0 newtons, 20 newtons or 40 newtons?


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Originally posted by memento
Did I make an error? Quite possible of course, where?

Oh I see what you were doing, you were adding up each x and y component of all the forces. You’re not supposed to do that.

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So simple stretch:

You pull on your penis with a 10 newton force, because there is no acceleration we have an equal and opposite component from the base pulling in the opposite direction with a 10 newton force. Is the tension 0, 10 or 20 newtons?

Right. The tension is 10N.

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With the 90 degree example you are pulling on your penis with a 10 newton force, because the Fulcrum is exerting an opposite effect it must be opposing this force at least equal to H. Its a reasonable assumption that P is also 10 newtons because the penis is not slipping over the fulcrum point. Therefore F is 20 newtons. So is the tension 0 newtons, 20 newtons or 40 newtons?

*grin* The tension is still 10N.

Take a look at the piece of penis you’re pulling with your hand. Imagine it’s chopped off just below your grip, if you will. You’re pulling up on it with a force of H. So in order for the piece of penis to remain stationary and not accelerate, there needs to be another force pulling in the opposite direction you are, with equal magnitude. This force that’s playing tug of war with you is the tension. Every single little piece of penis is being pulled apart by two opposite forces of that magnitude. And that magnitude, going up a few sentences, must be equal to the force you’re pulling with. Otherwise the little piece of penis you’re holding would accelerate. This always remains true; changing fulcrum, changing angle of fulcrum etc. do not change this one iota.

OK I see a problem with my thinking but its not the problem you describe, or at least not in the way you put it.

Going back to the 90 degree example again (effectively the glans meeting the base) The downward force may be 20 newtons and the upward force may be 20 newtons but effectively you are acting on double the penis cross sectional area.

So altering the angle is going to alter the effective diameter the force is working on down to a minimal cross sectional area with no fulcrum.

I think I have to do some more reading on this, which means pulling out my old physics books (damn).


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Ok, so is it pointless to do V-Stretches then?

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Ok, so is it pointless to do V-Stretches then?

I’m not sure. There may be some benefit to changing the shape or direction or something. But I am sure the overall tension in the penis is equal to the strength with which you pull, and in that a V-stretch is exactly the same as a normal stretch.

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OK I see a problem with my thinking but its not the problem you describe, or at least not in the way you put it.

Going back to the 90 degree example again (effectively the glans meeting the base) The downward force may be 20 newtons and the upward force may be 20 newtons but effectively you are acting on double the penis cross sectional area.

So altering the angle is going to alter the effective diameter the force is working on down to a minimal cross sectional area with no fulcrum.

I think I have to do some more reading on this, which means pulling out my old physics books (damn).

No… if you pretend the folded over penis is one object, then pulling up there are two 10N forces, and pulling down there is the single 20N fulcrum force. This makes a 20N tension, over double the cross sectional area. That’s exactly the same as a 10N tension over the normal cross sectional area.

This mathematical debate is waaaay above my ‘paygrade’, but I am still interested to see what you and mem come up with.

busted bus said he has a really sound mathematical background. I’m sure he’d be willing to contribute if or when he sees the thread.

Nedd,

The fundamental flaw in your analysis is the assumption of static rigid body model, where you basically assume the system is just a rigid lever or two pieces of rigid levers with a joint. A more realistic analysis would be elastic-plastic finite/boundary element analysis (the model used to analyze stress distribution in cars, tires and space shuttles these days. And BTW, memento, you can’t find it in high school physics books :) ), where if you plot the stress distribution, you’ll see the hot spots (high stress areas) mostly at and around the areas where you apply the stress. It could be an interesting term project for college structural analysis courses: “Penile Stress Analysis using FBEM”. I can already visualize the wire-frame CAD model used and abused in almost every sci-fi movie.

V-stretch is very effective to apply localized stress to tunica and septum. It’s not just in your head when you can feel the localized stress (the signals collected by the local nerve endings) :)


Quizás, quizás, quizás...

Nedd,

>No… if you pretend the folded over penis is one object, then pulling up there are two 10N forces, and pulling down there is the single 20N fulcrum force. This makes a 20N tension, over double the cross sectional area. That’s exactly the same as a 10N tension over the normal cross sectional area.<

Thats exactly the problem with my thinking, I ended up butting into some very simple physics there. As you increase the angle you increase the cross-sectional area, ending up with double, and yes halving the forces.

Every time I start thinking about this one I mean to actually follow through on research and every time I haven’t done so.

Cheers Tantrex,

My books are a little beyond High school - its just a long time ago. I never went to high school btw.

You’re right any model that assumes no elasticity is only of a certain worth and you’re also right that a full understanding is probably beyond me, but then its probably beyond you and Nedd, even if you happen to design suspension bridges or aeroplanes for a living. In the end all we can to is add together understanding and head toward approximations based on models, inserting different models when appropriate.

Nice points about stress distribution, which was effectively what Piet was saying earlier when I was trying to talk about the more general forces of MagnumXL’s model, but don’t you think we should be starting with Newtonian physics (and maybe contemporaneous physics .. Hooke?) and working up.

If all thats left at the end of that is that all the forces balance out and that the only advantage is localised stress, brilliant but I’d love to know why.

Personally I’m very interested in this, I used the V-stretch for most of my time PE’ing and I’ve been very aware that using it allows me to exert more force through the hand because I can stretch straight armed and used the power of the shoulder (or legs).


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tantrex - You’d probably need specialist measuring equipment to find out exactly what’s going on. And the analysis would doubtless be beyond any of us here.

I admit it’s possible the fulcrum could change the stress distribution. The magnitude of the average stress over any cross sectional area will be the same, but the distribution within that cross sectional area could be different.

What my analysis does do, however, is show that there is no magnification in average stress caused by adding a fulcrum. The small amount of deformation would have a negligible effect. Forces must still sum to (very close to) zero.

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