Question about Sizemisters data page

Guys,

I was asleep in math/science class…..

I was curious as to how volume is calculated…. I want to figure out what final size I will need to double the original size of my dick.

2*pi*the radius of your dick (root of [girth/pi]) * length

Of course, your dick is probably not a perfect cylinder, so it’s only an approximation.

Radius squared times pi times length
Pi is approximately 3.14

Divide your girth (circumference) by pi to find the diameter. The radius is half the diameter.

So, for a 7”x5” dick:

5/3.14 = diameter of 1.60”, radius = .8”

(.8)(.8)(3.14)(7) = 14.07 cubic inches

Our dicks aren’t cylinders, so this is only a rough estimate of the true volume.

woops, I said that radius was root of girth/pi.

As hobby wrote, it’s half of girth/pi

OK so right now my dick is 6.5-63/4-7 x 5….I’ll err on the smaller side at 6.5 x 5 I get a total volume of 32.5….is that right?

So to double my volume of my erect size how can I work backwards and see what my goal would be?

I tried this a couple of times .I got a large glass jar ,made a hole in the side near the bottom ,inserted a 24” long plastic hose and sealed around where they joined..[Just like a large vac chamber]

I filled the jar with warm water and squatted over it inserting both my dick and balls,Then pushing down hard to make a good seal I sucked out some of the water ,which in-turn made my dick and balls bigger .making sure no air got in .. Then I released the vac and took out my dick and balls .I then carefully measured how much water it took to refill the jar to the top. The first time it was just over 900 cc the second time 950 cc,,Which is just under one liter

6.5x5 comes out to 12.94 cubic inches. There are a jillion combinations of length and girth that would occupy 26 cubic inches. Pick either length or girth and figure out how big the other has to be to provide that volume.

For this example, let’s figure what girth is needed for a 9” dick to be 26 cubic inches:

Divide volume by length and you get the cross-sectional area.

26/9 = 2.89 inches. Now we have the area of the circle, we need to figure its circumference.

The area of a circle is pi(r^2). Divide the area by pi, and you get the radius squared.

2.89/3.14 = .920 = r^2

Taking the square root of this gives you the radius, .960

The circumference of a circle is pi(d). Double the radius to get the diameter.

(.960)(2) = 1.92.

Now multiply the diameter by pi to get the circumference.

(1.92)(3.14) = 6.03.

So, 9x6 is roughly 26 cubic inches. You can figure out the girth needed for other lengths.

Thanks alot man….I was actually shooting for 9 x 6…….just making sure….

No problem, BBS. 9x6 sounds good to me too.

Alright, I had too much time on my hands and worked through some examples here…

In the case of 6.5x5 (assuming 6.5 is length and 5 is circumference (aka girth))…

(Circumference / pi = Diameter)

5 / pi = 1.592

(Diameter / 2 = Radius)

1.59 / 2 = 0.796

(Estimated Volume = pi * Radius squared * Length)

pi * (0.796)^2 * 6.5 = 12.931 in^3

To double your volume, and assuming equal amounts of length and girth gain, you would do the following…

Double volume = 12.931 * 2 = 25.863 in^3

Now, I did a lot of algebra and had to solve a cubic equation, and got this…

New Length = 6.410 in
New Girth = 7.910 in

(For reference, my original cubic equation was as follows…

pi * [(5 + x) / (2 * pi)]^2 * (6.5 + x) = 25.863

which then reduces to…

x^3 + 16.5x^2 + 90x - 162.5 = 0

This gives a real solution of x = 1.410, or complex conjugate solutions of -8.955 + 5.921i and -8.955 - 5.921i, but since none of us have imaginary dicks, those two complex conjugate answers don’t apply, hence x = 1.140.

So, to get new length and girth, merely add 1.140 to the length and girths of 6.5 in and 5 in, respectively.)

However, if you wanted to grow in proportion to the current length and girth and wanted to double your volume, the following calculation is needed…

The current ratio of length to girth is 6.5 / 5 = 1.3

So after some drawn out algebra and solving for the only real solution to the cubic equation, you get the following…

New Length = 8.189 in
New Girth = 6.300 in

(For reference, my original cubic equation was as follows…

pi * [(5 + x) / (2 * pi)]^2 * (6.5 + 1.3x) = 25.863

which then reduces to…

1.3x^3 + 19.5x^2 + 97.5x - 162.5 = 0

This gives a real solution of x = 1.300, or complex conjugate solutions of -8.150 + 5.456i and -8.150 - 5.456i, but since none of us have imaginary dicks, those two complex conjugate answers don’t apply, hence x = 1.300.

So, to get new length and girth, merely add 1.300 to the girth and 1.689 (1.3 *1.300 = 1.689) to the length.)

Alright, I went kinda crazy with answering this question, but I figured I’d put my electrical engineering degree and my math minor to some use, hehe.

The best way to find volume however, would probably be the water displacement method as mentioned above. However, the volume estimation method is good as the estimate it is.

Oops, I messed up on the first example of doubling volume and put length for girth and girth for length. It should read…

New Length = 7.910 in
New Girth = 6.410 in

I was bound to mess something up on that reply. It figures that it was something pretty stupid. Oh well.

Thanks again for the answers…..but

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

dyslexic_smile wrote:

However, if you wanted to grow in proportion to the current length and girth and wanted to double your volume, the following calculation is needed…

The current ratio of length to girth is 6.5 / 5 = 1.3

Is 6.5/5 really the ratio? Is length/circumference an accurate model, or is length/area more appropriate?

I’m not trying to be a smartass, but I’ve wondered about this myself. If area is used for the girth number in the L/G ratio, emphasis is placed on r^2, which makes girth more of a factor. If circumference is used, as in your example, the squaring doesn’t come into play, so girth carries “less weight.”

Thoughts?

Lets get out the old differential equations now.

Related Rates and what not, make it real complicated ;)

Good thread to drum up bizz for Size’s PE data site. Click on the link at the bottum of the page, enter your data and it will compute your volume for you.