[Reason]

Thanks very much for clarifying.

The bottom of the head to base of the collar is usually quoted as being roughly a third of a head. So according to your measurements, a head should be about 13.5% relative to the height of the body, on average. Does this agree with your measurements?

And in case you were interested, that central hole at the base of the neck is the manubrium.

[stmercy2020]

And in case you were interested, that central hole at the base of the neck is the manubrium.

Point of interest, actually- the bone underlying that hole is the manubrium.  That hole is the suprasternal notch, which I've incorrectly called the subclavicle notch ever since being introduced to it as a control point in martial arts training.

Interesting uses of trig… I knew there was a reason to study it in high school. 😀

[John]

a head should be about 13.5% relative to the height of the body, on average.

That's about 1/7, which I've gathered is the head/height ratio for most humans.  It does seem this would be higher for shorter people, as heads generally don't change sizes that much between tall and short adults.

[1st_Tsurugi]

I'm a math midget.  I'm not entirely certain I understand exactly how your ratios work.  Is there a chance you could give it to me in other words?

[John]

I'm a math midget.  I'm not entirely certain I understand exactly how your ratios work.  Is there a chance you could give it to me in other words?

Well for the width-to-depth ratios, let's say you take a limb 10 inches deep. A 0.7 ratio would be 7 inches wide for every 10 inches of depth, so width = 7" and depth = 10".  If the image is taken from somewhere between a width cross-section, and a height cross-section, then the image's cross-section's length would be somewhere between width and depth.  The formula I gave assumes the cross-section's length goes steadily from width (0 degrees) to depth (90 degrees).  So if the rotation is estimated at 22.5 degrees (1/4 of 90 degrees) rotation, the length of the cross-section would be

7+(10-7)*(22.5/90)
=7+3*1/4
=7+0.75
=7.75"

So if you measure about 7.75 scale inches on the photo or art, finding width and depth would be like this

Ratio (r):
r=0.7

Intermediary between r and 1, corresponding to the image's cross-section (t):
t=r+(A/90)*(1-r)
=0.7+(22.5/90)*(1-0.7)
=0.7+0.25*0.3
=0.7+0.075
=0.775

Measurement (m):
m=7.75"

Depth (d):
d=m/t
=7.75/0.775
=10"

Width (w):
w=d*r
=10*0.7
=7"

a & b radii for ellipse:
a=w/2=7/2=3.5
b=d/2=10/2=5

pi Approximation:
pi=3.14

Approximate Circumference (C):
C=sqrt(2*(a^2+b^2))*pi*r
=sqrt(2*(5^2+3.5^2))*3.14*0.7
=sqrt(2*(25+12.25))*2.198
=sqrt(2*37.25)*2.198
=sqrt(77.25)*2.198
=8.789*2.198
=19.318"

—-

For the height ratios, it's just % of total height.  You either need to guess what your character's height is, or you can use the real height of the person you are morphing or otherwise basing the art off of.

[1st_Tsurugi]

I think I understand most of the individual things you laid out, but my math deficient mind can't quite put it all together as yet.  I guess it'll make sense later.  Thanks for trying 🙂

[JimmyDimples]

Thanks very much for helping crunch the numbers, and show the terms.  While I don't draw this stuff, it's good to know when I put the physique into words.  🙂

[John]

Ok, now I'm trying volume and weight.  I ran a test on a very muscular FBB, and got the following percentages.

[font=Courier New]PART (Shape x (copies)) ..% of volume (all copies)
THIGH (Ellipsoid x 2) ….21.1%
ANKLE (Frustum x 2) ……1.8%
CALF (Ellipsoid x 2) …..9.2%
BUTT (Ellipsoid x 2) …..9.7%
ABDOM (Frustum x 1) ……14.2%
CHEST (Cylinder x 1) …..28.4%
TRAP (Ellipsoid x 0.5) …0.6%
NECK (Cylinder x 1) ……0.9%
BICEP (Ellipsoid x 2) ….6.0%
FOREARM (Ellipsoid x 2) ..4.7%
DELT (Ellipsoid x 1) …..3.3%[/font]

I'm leaving out Head, Feet, and Hands as they don't differ much from person to person in weight or size.  I may need to get a common skeletal weight to figure in for weight.  Do most of these volume proportions look correct, for anyone familiar with the subject?

The bodybuilder in question was 5-6 (66 in), and photographed at about 180 lbs.  Her "muscle volume" came to 4174 cubic inches.  In other words, if she was redistributed into a cube, all of the cube's dimensions would be 16".

[John]

I may not have perfected the method (as one could imagine it is pretty hard to get exact off just a picture), but I have put together a spreadsheet to be used for approximating weight off a picture.  Far from an exact science, it at least beats trying to guess the weight just by eyeballing.  Right now it is hosted on my Yahoo! group (you have to sign up for the group).  If anyone wants to host it elsewhere, be my guest.

Weight Measurement Aides

Here is an illustration of the volume zones used.

Of course, part of the fun of drawing amazons is guessing the weight.  Even so, I grew tired of just guessing and wanted a better method.  My guess have been pretty accurate for the two fantasy girls I tested out.  I also tested it on a few real FBBs, and the weights seemed to hold up.  With the inexact method, there will still be some guesswork involved, but now you can at least have a "neighborhood" to place your guess in.

[trilliwig]

Hi MuskelGrothe.  Many thanks for creating this useful tool!  I've been trying to apply it to a commissioned piece of artwork, and run into a couple issues for which I have questions.  ???

First, the example spreadsheet entries for the chest appear to use a width greater than the length (75 pixels versus 45 pixels).  Your diagram shows chest width as the measurement from collarbone to abdomen, while the chest length is the measurement across from side-to-side, so by the diagram, length should be greater than width.  Which is correct, or does it matter?

Second, your ratios width/depth for thighs, biceps, and calves don't give guidance for a normal, non-muscular woman.  I tend to use the ratios closest to 1 in that case.

Third, the weight sheet contains an input for Target weight in pounds.  Isn't that what we're trying to calculate?  So what purpose does that serve?  For now I've been putting in something that approximates 0.036 lb/in^3, erring on the low side for less muscular women and on the high side for more muscular women.

Also, I've put in a row for elliptical circumference using Ramanujan's approximation, which is

pi*[3(a+b) – sqrt((3a+b)(a+3b))]

where a and b are the major and minor radii.  This seems to work out pretty well.  An example formula is

=PI*(1.5*(C8+C9)-0.5*SQRT((3*C8+C9)*(3*C9+C8)))*$B18/$B17

which gives a circumference in inches.

[Author]

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