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Gain potential with pumping after 1 inch gain with manual exercise's?

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Gain potential with pumping after 1 inch gain with manual exercise's?

In all the time I have been Peíng I think I have gained at least 1 inch, 1,18 max (3 cm)

At this forum I gained .5/6 inch.

The rest a year before, when I downloaded a pe document at kazaa. Was not aware enough about it and forgot to count it in at this board for myself.

Now after this said, what would my potential of pumping be?? (after the gains I already have)

I am 6.7 inch now.

Could I get to a solid 7.5 inch in 22 months with pumping/pe weights/a-e/homedic wrist wrap??

I, can, I can, I can not, can not compute..

Dunno, anon. Everybody’s different and because of that our crystal balls are foggy.

You won’t know unless you try, so go for it.

_______________

avocet8

Thanks A8,

I have a thought/feeling about pumping.

Thinking that you can always gain with it, because you can literally see you’re dick growing in the tube.

Assuming the growth in the tube will translate in erect growth…

I, can, I can, I can not, can not compute..

There is a formula for calculating your potential gains. Here it is:

A(0) / 2 + (k=1..) [ A(k) cos (k(PI)x / m) + B(k) (sin k(PI)x / m) ]
a(k) = 1/m f(x) cos (k(PI)x / m) dx
b(k) = 1/m f(x) sin (k(PI)x / m) dx

You do realize that the universe will end in 49 minutes now.

This place runs on donations, help out if you can. Thanks.

Originally Posted by gprent
There is a formula for calculating your potential gains. Here it is:

A(0) / 2 + (k=1..) [ A(k) cos (k(PI)x / m) + B(k) (sin k(PI)x / m) ]
a(k) = 1/m f(x) cos (k(PI)x / m) dx
b(k) = 1/m f(x) sin (k(PI)x / m) dx

My potential gains are 0.3643”, which I have already surpassed, so your formula is flawed.

As for the threadstarter, I say go for it! You have nothing to lose!

Current PE status - Contemplating Retirement. STARTED - 6.75"x5.25" CURRENTLY 7.5"x5.5" - BPFSL - 7.25"

Originally Posted by gprent
There is a formula for calculating your potential gains. Here it is:

A(0) / 2 + (k=1.) [ A(k) cos (k(PI)x / m) + B(k) (sin k(PI)x / m) ]
A(k) = 1/m f(x) cos (k(PI)x / m) dx
B(k) = 1/m f(x) sin (k(PI)x / m) dx

:rofl:

You do realize that the universe will end in 49 minutes now.

Double :rofl:

Why worry about what your potential is? Maybe you’ll get there, maybe not- if we tell you that 7.2 is your max potential, will you quit PE? Of course not, so set a goal, and strive onward- but don’t let the goal interfere with your PE.

Remember, you are also getting a much healthier peepee

WE are the 99% 'WE are the people you depend on; we cook your meals, we haul your trash, we connect your calls. We drive your ambulances. We guard you while you sleep. Don't f&ck with us'-- Madame DeFarge

"Rope trades @\$10 a yard. I wonder if they even know that?"- Capitalist

Last edited by androNYC : 09-01-2005 at .

Originally Posted by gprent
There is a formula for calculating your potential gains. Here it is:

A(0) / 2 + (k=1..) [ A(k) cos (k(PI)x / m) + B(k) (sin k(PI)x / m) ]
a(k) = 1/m f(x) cos (k(PI)x / m) dx
b(k) = 1/m f(x) sin (k(PI)x / m) dx

See if you can figure this one out:

The farmer wants to get his goat, wolf and cabbage to the other side of the river. His boat isn’t very big and can only carry him and either his goat, his wolf or his cabbage. Now…..if he leaves the goat alone with the cabbage, the goat will gobble up the cabbage. If he leaves the wolf alone with the goat, the wolf will gobble up the goat. When the farmer is present, the goat and cabbage are safe from being gobbled up by their predators.

How does the farmer manage to get everything safely to the other side of the river?

It may not help you figure out potential gains, but if you can answer it you may not need a formula.

Originally Posted by travelguy14
The farmer wants to get his goat, wolf and cabbage to the other side of the river. His boat isn’t very big and can only carry him and either his goat, his wolf or his cabbage. Now…..if he leaves the goat alone with the cabbage, the goat will gobble up the cabbage. If he leaves the wolf alone with the goat, the wolf will gobble up the goat. When the farmer is present, the goat and cabbage are safe from being gobbled up by their predators.

How does the farmer manage to get everything safely to the other side of the river?

Take the goat across.
Take the cabbage across.
Take the goat back to the original side leaving only the cabbage.
Take the wolf across leaving the wolf and cabbage on the other side.
Go back and get the goat and take him across.

:)

Cut them all up for a nice stew, eat your fill, then worry about getting to the other side of the river.

This place runs on donations, help out if you can. Thanks.

Stuff the cabbage with goat wolf, thats what I’d do, then take them over in your belly.

Good suggestions but not correct yet. I like the stew idea, LOL.

This one is easy. He leaves the boat and walks across the bridge with the goat, wolf and cabbage. :)

:buttrock: The Peter Dick method :buttrock:

Then, BPEL:7.500"x5.500"

Now, BPEL:8.375"X6.750"

Originally Posted by travelguy14
Good suggestions but not correct yet. I like the stew idea, LOL.

I could of swore Toolguy had it. No?

OK, Here is the solution:

The farmer must first take the goat across the river. He then returns and picks up the wolf. He leaves the wolf off and takes the goat back across the river with him. Then he leaves the goat at the starting poing and takes the cabbage over to where the wolf is. He returns and picks up the goat, and then lands where the wolf and the cabbage are!

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