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scotch

Rough_Rock
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I was trying to calculate the exact dimensions including crown and pavilion depths from the standard dimensions as given on a GIA cert, namely spread (diameter), table, total depth and carat weight. So I approximated the total volume of a round brilliant as the composition of a cone (for the pavilion), a truncated cone (crown) and a flat cylinder (girdle). Using some highschool math as well as the density of a diamond (3.51 g/cubic cm), one should indeed be able to calculate crown and pavilion depths from the standard measurements. However, I found the formula not very useful, apparently for two reasons:

(1) it seems I have to assume a much larger girdle than given on the cert, e.g. if the girdle is "medium", say 1.3%, I have to use 2.6% to arrive at the correct values. To put it another way, pavil depth, crown depth and girdle don't add up to total depth.

(2) the calculation is extremely sensitive to changes in girdle width, e.g. a 0.1% variation in the girdle can gives many times that variation in crown depth.

Does anybody know how to do this correctly?
 

Richard Sherwood

Ideal_Rock
Joined
Sep 25, 2002
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4,924
Very interesting, Scotch. Let me see if I'm understanding you correctly. Are you saying that because the carat weight, total depth, table size and specific gravity are available, you can (roughly) determine the crown height and pavilion depth by means of the "volume" of a diamond?

In regards to your question -
-----------------------------
>>>>>it seems I have to assume a much larger girdle than given on the cert, e.g. if the girdle is "medium", say 1.3%, I have to use 2.6% to arrive at the correct values. To put it another way, pavil depth, crown depth and girdle don't add up to total depth.-----------------------------

I was asking this question myself not too long ago. Garry answered it correctly with the observation the girdle thickness derived by Western gemologists is determined by measuring the thinnest and thickest portion of the girdle at the "valleys", rather than the thinnest and thickest "high points", or even the thinnest "valley" and the thickest "high point" (which really would give the most correct total depth measurement, IF the crown height and pavilion depth were measured as starting from the same "average" point, instead of from the "high points", which is how they are measured).

Using those measurements (thinnest to thickest "valley"), the total depth will always come up short (as you observed), usually in the neighborhood of 1.3 to 1.7%.

-----------------------------
>>>>>the calculation is extremely sensitive to changes in girdle width, e.g. a 0.1% variation in the girdle can gives many times that variation in crown depth.-----------------------------

Perhaps that is related to the fact discussed above. Since the measurement being used is smaller than it actually is (1/2, or thereabouts), the variation creates a comparable greater variation in crown height or pavilion depth?

One website you might find relevant to this is Marty Haske's website ADAMAS. He has an article on girdle thickness which is quite fascinating. I'd be interested in your opinion on it.

Rich
 

Garry H (Cut Nut)

Super_Ideal_Rock
Trade
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Aug 15, 2000
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18,418
Scotch why are you interested in diamonds?
Surely you have not one romantic bone in your analytic body? :)

Seriuosly though you need to download DiamCalc (it is listed on the bar at the top). Play with the free marquise and buy the product if you like it. I think you will :)

Because the girdle is the largest cross sectional part of the diamond is is critical to weight, so it is correct that you have big variances. And Rich is correct in all he has told you :)
 

scotch

Rough_Rock
Joined
Oct 15, 2002
Messages
94
Thanks for your reply, that was really helpful. :wavey:
----------------
> Very interesting, Scotch. Let me see if I'm understanding you correctly. ....?

-----------------------------
Yes, exactly. If you approximate the shape of a round brilliant by simple geometric shapes such as cones, truncated cones and cylinders, if you further assume that girdle (measured the correct way), pavilion and crown depths add up to total depth, multiply by specific weight and convert to carat, you arrive at an equation that you can resolve for the unknown variable (either crown or pavilion depth): (Vol(cone) + Vol(cylinder) + Vol(truncated cone))*density/0.2 = carat weight. You only need to insert the Volume formulae, e.g. Vol(cone) = 1/3* (Base diameter/2)^2 *(height).
-----------------------------
> I was asking this question myself not too long ago. ....

-----------------------------
Makes total sense to me, thanks a lot.
-----------------------------
> Perhaps that is related to the fact discussed above. ...

------------------------------
Your right, if the stated girdle size is about 1/2 the "real" one, that means you have a double effect of a change in girdle size right there. I'll check out the ADAMAS site and get back to you.

Yours,
Scotch
 

scotch

Rough_Rock
Joined
Oct 15, 2002
Messages
94
--------
> Scotch why are you interested in diamonds?
Surely you have not one romantic bone in your analytic body? :)
--------
Let's see how romantic it gets when she sees the stone I've finally selected...:twirl:
------

> Seriuosly though you need to download DiamCalc (it is listed on the bar at the top). Play with the free marquise and buy the product if you like it. I think you will :)

-------
I have actually downloaded the trial version already. However, I'm looking for a diamond only for one special occasion, so I don't think it would make sense for me to buy it. Don't get me wrong, I don't underestimate how much knowledge and research went into it, and it's certainly worth the price.

By the way, the HCA website was extremely helpful to me. I'm waiting for my Idealscope to arrive...

Best regards,
Scotch
 

Richard Sherwood

Ideal_Rock
Joined
Sep 25, 2002
Messages
4,924
Scotch, does that formula allow for varying crown heights and pavilion depths? I mean, for a 60% total depth stone you could have cr.ht./pv.dp. rations such as 14%/43% or 8%/49%.

Are you able to figure out what the particular crown height - pavilion depth is for particular stones?

Rich, was playing hookey during that class...
 

Garry H (Cut Nut)

Super_Ideal_Rock
Trade
Joined
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Messages
18,418
Rich you are really going to need to buy DiamCalc.

The answer to your question is all there and takes a few seconds to calculate - the really deep pavilion stone weighs 4%less. I bet you would have thought diferently - I would have.
 

Richard Sherwood

Ideal_Rock
Joined
Sep 25, 2002
Messages
4,924
Yeah, the DiamondCalc is on my list. I was talking to Leonid about it.

It seems like it might be a good tool in my appraisal reports. What do you think? Show the "theoretical" light performance of a client's diamond, along with how it compares to a "scale" of light return "theoretical" photos from various other cut quality grades.

Also, let me know when you get the light box on the market for your IdealScope. Will it aid in taking photos of the IdealScope image?

Rich
 

pricescope

Ideal_Rock
Joined
Dec 31, 1999
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8,266
Richard, I replied to you regarding DiamCalc but there were problems with our email server - not sure whether you received it.
 

scotch

Rough_Rock
Joined
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Messages
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*** Hi Webmaster, how do I type plus and minus signs in here??? ***

Rich, here is the complete formula:

(1) (Pi/4*(1/3*D^2*pl plus cb/3*(t^2 plus t*D plus D^2) plus D^2*gb)*0.00351g/mm^3)/(0.2ct/g) = W
(2) pl plus cb plus gb = H

I've adapted the Notation from DiamCalc:

D = (average) diameter
H = Height (total depth)
pl = pavillion height measured at lower facets
gb = girdle width measured at bezel facets
cb = crown height measured at bezel facets
t = table
(all in mm units)
W = weight in carats
Pi = 3.14159265359...

You can put formula (1) in a spreadsheet, and using (2) to replace pl by (H minus cb minus gb) you can solve (1) for cb, or you can replace cb by an expression containing pl you can solve for pl (using excel solver, if you like). The formula works only if you have no culet, if there is a noticable culet, you would have to replace the term 1/3*D^2*pl by a volume formula for a truncated cone, similar to the second term in (1).

Best regards,
Scotch
 

Richard Sherwood

Ideal_Rock
Joined
Sep 25, 2002
Messages
4,924
Awesome Scotch, thanks for the formula.

Leonid, I did get your email, and downloaded the sample DiamondCalc program with the marquise. Does buying the program just involve getting a registration number from you that "unlocks" the program?

Awesome looking program.

Rich, GG
Sarasota Gemological Laboratory
 

pricescope

Ideal_Rock
Joined
Dec 31, 1999
Messages
8,266
Scotch,


----------------
*** Hi Webmaster, how do I type plus and minus signs in here??? ***
----------------


Alt+0177: ±

:)
 

pricescope

Ideal_Rock
Joined
Dec 31, 1999
Messages
8,266
1. send $280 via paypal,
2. email me your registration code,
3. developers will email you registration keys the same or next day.

I will be happy to answer all your questions on using the program. How about phone chat?

:)
[/u]
 

scotch

Rough_Rock
Joined
Oct 15, 2002
Messages
94
My pleasure. btw, you may want to change the density from 0.00351 to 0.003522, that's the value DiamCalc is using. Makes a big difference.

Best Regards,
Scotch
 
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