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Originally posted by memento
>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.<OK here we go. So using that as a starting place that would mean you agree there is a 20N force. It would also mean that at the top of the loop there is a 20N force working over the single cross sectional area. So this is our local tension increase.
Huh? At the joint/fulcrum, there’s a 20N force, but it’s not working over a single cross sectional area adding to the tension like the 10N forces are. This 20N force only acts at the joint, and even there it’s not creating tension. At the joint things get more complicated… I left out detailed analysis of the joint because of that. Frankly I don’t know what goes on at the joint. Like I said it’s complicated, you’d probably need sophisticated measuring equipment to know exactly what’s going on. My guess is there’d be some compressive stresses in there and maybe even some reduced overall tension.
Okay, here we go. I could very well be wrong with this, I’m just making it up as I go along, but: Imagine a cable going round a pulley. On a section of cable around that pulley area there are two tension forces tangent to each end of the arc formed by the section we selected. There is also a radial force exerted by the pulley. This would suggest that there is reduced tension and added compression; the x components of the tangential forces produce tension, and the y components of the tangential forces plus the radial force produce compression. But. Take an infinitesimal piece, and the two tangent forces are parallel. Thus the radial force must be zero, as makes sense considering the fact that there is only an infinitesimal area exposed to the stress the pulley exerts on the cable. Interesting.
So what does that mean? It means, an ideal cable making a continuous curve still has zero compressive forces acting on it and the same tension as any straight section of cable.
How about a discontinuous curve? I guess the points along the discontinuous joint have reduced tension but added compression.
The penis isn’t discontinuous though. It’s also not ideal. So really, I don’t know what exactly’s going on inside the penis. Again, you’d very probably need sophisticated measuring equipment to know.
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Originally posted by memento
So if we have a simple 20N force over under a thumb and a 10N force at the extremities it seems reasonable to assume that the tension gradient will actually cause effect over the whole shaft. We are directly effecting tension in a fair percentage of the penis (somewhere between 10-20%), so the overall tension in the shaft should increase.
No. The second a tension gradient exists within a cable, it snaps. Reductio ad absurdum, your reasoning is flawed. I direct you to my first post yet again. The tension in the straight section of penis on either side of the joint must be uniform. Must be. Any reasoning you come up with that shows otherwise is flawed. Even when the penis is not an ideal cable, this remains true.
I don’t know what exactly goes on around the joint, because around there the penis is no longer a simple cable. But I don’t have to know anything about the joint to know that the tension in the straight section of the penis leading up to the joint is uniform, and equal in magnitude to the force you’re pulling it with.