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# Question about Sizemisters data page

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Alright, my answers are any of infinite possibilities for doubling the volume of a cylinder with a length of 6.5” and circumference of 5”. I was merely pointing out two distinct possibilities. The first case is assuming that the length and girth both increased by an equal amount. The second case was the case in which the ratio between length and circumference was kept constant.

Penises come in many different sizes, but I think that 6.5x5 is a very common size, so the L/G ratio of 4/3 is what I looked at. This will keep the general size of the penis the same, but with simply an enlargement factor applied to it. For instance, I came up with 8.189x6.300 as a doubling of volume, while keeping the L/G ratio intact. In this case, both the length and the girth are 1.260 times bigger than they were in the original case of 6.5x5 (a gain in size of about 26%).

I suppose you could examine looking at length vs. area, but I’m not sure how that would work exactly. Area is a function of the cirumference ( Area = [(Circumference)^2 ] / (4 * pi) ), so it does get taken care of properly I believe. I think it’s just easier to compare length and circumference (or width even), simply because that is what is used in PE terminology the most often. So you could find a correlation between length and area and find a formula to solve for the proper “ratio” between them, although I think this could be a bit tricky. This is mostly because you are comparing units of inches to units of square inches, and therefore have that whole squaring issue to worry about.

hobby, I guess I see your point about comparing the area, but in my opinion, I think the area gets taken care of properly as long as you keep the ratio between current length and current circumference intact. That way, the penis has the same general shape, but is simply a bigger version of itself.

In my case (6.5x5 to 8.189x6.300), the width of the penis also increases by a 26% factor, as diameter increases linearly with circumference. So, if you were looking down at your own erect penis while standing up (looking at it as a 2-dimensional object), its width and length will have increased proportionally, so therefore it would look like the same “rectangle” (I know I’m making a big stretch there in describing the shape), only the length and widths will have increased by the same factor, hence keeping the same aspect ratio.

If you also took a cross-section of the penis in the case I mentioned, the area of a cross section of the penis would increase by a factor of 1.587, or about a 59% increase in actual area (1.989 square inches vs. 3.158 square inches). This seems like a significant increase to me, as yes, this is more than double the increase of the 26% for the cases of width or circumference or length (due to the squaring action).

So, if you were to flop things around, and only had a 26% increase in penis area, that would only result in a little over a 12% gain in girth. And I kinda just forgot where I was going with this argument, so I’ll stop for now, haha.

Alright, I guess it’s apparent that I’m getting tired here, so I’m going to try to get a quick nap in before I have to be up in a couple of hours. Not sure if I was answering any questions along the way, or just merely complicating matters, but I hope anything I said made sense.

In any of my visualizations, it does help to picture the penis as a cylinder, which becomes a rectangle if you were to imagine it in a 2-d plane, assuming you were observing it from its upper surface (like standing up and looking down at your erect penis). As for the cross-sectional view, assume basically that you are looking head on at the penis, and if you were to take a slice (ouch!), this would be in the shape of a circle. I guess it helps if you ever took cad or know anything about orthographic drawings. I would use more anatomical terms to describe the various planes and surfaces, but I don’t think they would be of much help in this case.

Anyways, I’m rambling on, so I am cutting myself off now.

Quote
Originally posted by bigblackstick

which one is right or are they both right? is there infinite possiblities?

We are both right, and there are infinite combinations of length and girth that will give 26 cubic inches. A 3” dick would have to be 10.43 inches around! My question is about proportion. If you increase the size of a dick of x size y percent, what would it measure?

<plug> I agree with iamaru that everyone should be using the database. </plug>

dyslexic_smile,

My math brain cells are tired. They aren’t used to the level of exercise they had when I was in school. So, please cut me some slack if I manage to stick my foot in my mouth. :)

You wrote:

Quote
In my case (6.5x5 to 8.189x6.300), the width of the penis also increases by a 26% factor, as diameter increases linearly with circumference. So, if you were looking down at your own erect penis while standing up (looking at it as a 2-dimensional object), its width and length will have increased proportionally, so therefore it would look like the same “rectangle” (I know I'm making a big stretch there in describing the shape), only the length and widths will have increased by the same factor, hence keeping the same aspect ratio.

I can gain girth without my dick increasing in width as measured across the top. It can become less oval-shaped. Size’s computation of my width based on my girth does not reflect my true width. Mine’s wider. Penises aren’t exactly round, though when we attempt to compute volume we presume they are.

That said, actual width is irrelevant to my proportion question.

Here’s what I’ve thought:

Take a 6 x 5 dick and increase its size by 50%. What do you get?

l=6
r=.80

6(1.5) = 9 inches length

50% girth increase based on circumference = (1.6)(3.14)(1.5) = 7.54 inches

Ok, now do the same basing the girth increase on the cross section:

Initial area = 2.00 square inches. 50% increase = 2(1.5) = 3 inches

3 inches of area equates to 6.14” in circumference.

Which is the true 50% increase in size: 9 x 7.54 or 9 x 6.14?

It depends on whether you base the percent of girth increase on r or r^2. I think r^2 is probably the better measure. If you set up an equation bringing volume into the the mix, you’ll still have to decide whether to address girth based on circumference or cross section, so you can’t escape the issue.

Last edited by hobby : 02-20-2003 at .

hobby, after doing some number crunching I concluded the following…

If a penis went from 6x5 to 9x7.54, the volume increase would be around 225% (from 17.904 in^3 to 40.286 in^3).

If the penis went from 6x5 to 9x6.14, the volume increase would indeed be 50% (from 17.904 in^3 to 26.857 in^3).

So yes hobby, you are correct with the assumption of using the girth increase on the cross-section to get double the size (in volume). So, I admit I was wrong in this case.

I kind of thought that is what you meant when you were asking about area and what not, but I wasn’t sure, so I just went on my line of thought that I was on and rambled incoherently until I was ready to go to sleep, hehe.

Anyways, hope that solves this mystery. If there is further discussion, it may be a good 5 or 6 days until I reply, as I have to go out of town today (tomorrow in worst case). Either way, I hope that helps explain things better.

Hmm, I screwed up again…

When I said the first case was a 225% gain, I meant to type that it was a 125% gain. Either way, increasing girth based on cross-section will give the 50% gain.

Alright, I’m just heading out the door. Later all.

Perhaps we could have all these calcs redone in metrics [eg cubic centimeters and litres ]so the other 90% of the people, living in the third world countries, can understand what you are talking about.

Qualified Australian Stupidiologest

## Do them in Both

Please do them in both so we keep us Non-Americans happy. We don’t want another uprising or revolution in far off lands.

“You see, I don’t want to do good things, I want to do great things.” ~Alexander Joseph Luthor

I know Lewd Ferrigno personally.

Well the formulas are the same, just plug in your metric numbers.

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