FFFg Density (gr/in3)?


Steve Swartz
April 29, 2007, 03:11 PM
What is the density of Goex FFFg or where could I look it up?

I need to know how much powder (in grains) is needed to completely fill my "patent chamber" (0.06 in3).

So if I just knew what the grains per cubic inch were of black powder . . .

Steve in North Texas

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April 29, 2007, 03:26 PM
Steve, it varies by manufacturer and even by lot. You might try:


They have had discussions on density in the past. They'd be the folks to ask. I'm sure they'd get together and weigh some samples for you. They're a friendly and knowledgeable bunch. If you sign in with that link, I get a referrel credit, whatever that is. :)

April 29, 2007, 06:41 PM
See chart #2 for conversion info:


April 30, 2007, 08:51 AM
Does 777 provide the most power per volume? What about when compressed?

I'd like to find out which powder can provide the most power. I've also heard of people sifting it?

Steve Swartz
April 30, 2007, 10:52 AM
Thanks all.

After much finagling looks like my ~0.06 or so cubic inches is very close to a milliliter (makes sense- pedersoli gun!).

And it looks like a milliliter holds right around 15 gr of FFFg.

I crunched this several different ways with data from different websites and applied my trusty CRC manual . . . and it all came out fairly consistent.

O.K., so given that it takes around 15 gr of powder to fill my chamber . . .

. . . would I need a load of at least 15 gr for optimum consistency?

The mfgr recommends 12 gr load- but this is leaving an airspace over the powder and does not allow the ball to compress the load firmly.

Bad mojo?

Steve in North Texas

April 30, 2007, 12:05 PM
Are you talking 12 to 15 grains by weight or by volume? They are not the same.

You've determined that it takes 15 gr fffg (by volume?) to completely fill the "patent chamber". But the manufacturer recommends 12 gr fffg (by weight, perhaps?)

If the two numbers are both by volume then there may well indeed be an "airspace" using a 12 gr load; however, wouldn't the spherical shape of the ball fill that space and allow the ball to compress the powder? I assume the patent chamber is cylindrical but just a smaller diameter; if that's so, part of the ball (the majority of the lower hemisphere) will enter the empty space in the patent chamber until it rests on the powder. Correct?

Bad Flynch
April 30, 2007, 12:41 PM
>After much finagling looks like my ~0.06 or so cubic inches is very close to a milliliter <

1 Cu. In. = 16.387 cc. And that really is cc, not ml because ml is not a dimensionally derived volume quantity and the cc is.

15.432 grains = 1 Gram.

I took a look at the black powder conversion charts listed and that is a neat deal, but at least one of them is based on a faulty assumption. My Goex does not have 100grains by volume to equal 100 grains by weight. It varies, quite a bit, but the multiplication factor is about 0.85. So 100 grains by weight is around the equal of 100/0.85, or 117 grains by volume equivalent.

All of this fails to mention that your scooping technique can vary the fill for a single meaure quite a bit, but it is not significant usually if your technique is cnsistent.

Basically, the grains by volume measurement stems from an era when people did not have access to scales and, needed an easy way to measure powder in the field. It was derived by finding the volume of, say, 100 grains of water and then using that same volume of powder and calling it 100 grains by volume.

What you end up with is a sort of bulk measurement using specific gravity (bulk specific gravity). Most people equate specific gravity with density, but that is only a happy coincidence. Specific gravity is the weight of a substance compared with the weight of an equal volume of water. It just happens that density comes close, sometimes.

Density is measured in grams per ml. Grams is a measurement of mass, not weight, even though it is used that way. In theory, the amount of mass would not change with a move from here to the moon, even though the gravity factors are very much different. There is such a thing as a bulk density, and that has been used with powder. Specific gravity, however, is a weight-based measurement.

It is better to use either an all-weight system or an all-volume-type sytem. Going between the two can lead to confusion.

Steve Swartz
April 30, 2007, 01:03 PM
Never knew people used "Grains" to mean either weight or volume . . . sounds like a good "weigh" to confuse things!

I used density = weight (grains)/volume (in^3) which I believe is a pretty good working definition.

Started with baseline data developed experimentally - guess I have to trust the guy who reported this on his website (http://www.tbullock.com/bpsg.html):

********excerpt from website***********

Powder Density I made a cylindrical powder measure with a depth of 1.000 inch, a diameter of 1.128 inches, and a volume of one cubic inch. I measured and weighed all the powders I could get ahold of, and measured some power dippers.

Density of Black Powder
Powder Size Grains per
cubic inch Notes

GOEX Fg 235
GOEX FFg 231
GOEX Ctg 252 Surprisingly Dense
Superfine FFFFg 232 70 years old and good as new
Superfine FFFg 243
Kik FFg 230

242 Antique powder dipper
248 1970's T/C powder measure

Average 239

Quite a variation. That is why you measure powder by weight and not by volume. The average seems high to me; 235 is the value that I would likely use when making a powder measure.

********end excerpt from website***********

The volume of my patent chamber was estimated from a variety of cylinders using "go-no go" for diameter and depth of penetration. Couple of assumptions made (assumption of cylindrical shape, for one) but pretty sure my measurement is at least accurate, if imprecise.

Then calculated the volume using our old friend 3.14159 and pi*r^2*l to get volume of patent chamber.

Took density of black powder from table (above; used 238 gr/in^3) and multiplied by volume of patent chamber (cubic inches cancel out) to get grains in weight of the black powder.

I'm pretty sure I did all that right . . . I didn't realize that density or volume had different interpretations (but that mass vs. weight thing always gave me pause!).

Steve in North Texas

May 10, 2007, 12:20 PM
"1 Cu. In. = 16.387 cc. And that really is cc, not ml because ml is not a dimensionally derived volume quantity and the cc is."

Badflinch. I think you may have made a little mistake?
cc = cubic centimeter and ml = milliliter
1000cc = 1 liter
1000ml = 1 liter
These are exactly identical volume measurements (1/1000 of a liter) it's just a different name.

1 liter = 1000 grams (=1 kilogram) = 1000cc
1 cubic centimeter of water (cc) is therefore exactly 1 gram.

The metric system is very simple because volume, weights and distance (1000 centimeter = 1 meter and 1000 meter = 1 kilometer) are all very logically linked together by the thing we have most of on our planet. Water.

I still do not understand the volume grains however.
Is 1 grain by volume a certain volume in cc?
Or should you just multiply the grains by weight with 0.85 to get grains by volume? If so, the volume grains would vary with the different powder weights and volume weights would therefore not be consistent.
Am I Right?
This is confusing, or maybe I'm not smart enough:confused:

May 10, 2007, 01:25 PM
No way you couldn't just pour in a bit of powder slowly until the chamber is filled, then pour it out and measure with a powder measure how much it took? What kind of gun is it you've got?

Bad Flynch
May 10, 2007, 06:01 PM

>cc = cubic centimeter and ml = milliliter<

Depending on your definition date and source, the cc and the ml may or may not be exactly the same and have not always beeb derived the same way. I referenced the cc because it is the volume of one cubic centimetre of water, i.e., a cube 1 centimetre by 1 centimetre by one centimetre. The point being, is that the Cubic Inch is derived the same way, a cube of water 1 inch by 1 inch by 1 inch.

The ml has had more than one definiton over the years, and up until 1964 differed by about 27 or 28 parts per million from the cc. See Wikipedia for the explanation of the differences; they do it much better than can I (see Liter or Litre).

The history of the metric system has been long and complicated, with many changes in definition. At this late stage of the game, they only usually tell you the simple version.

The volume grain is best likened (as before) to a convenience measurement. It was conceived in a day when one could not actually weigh charges out. Can you imagine trying to weigh 65 grains of musket powder in the heat of battle? What one does is to weigh out the needed number of powder grains by weighing out the same number of grains of water and then using the same volume of powder as the water, and that is all there is to it. It is best described as using the bulk specific gravity as a measuring tool. It is simpler to do than to write about.

The volume grain can be converted to cc, but in a way it is a nonsense measurement. You can weigh out the number of grains of powder needed using water and then find the metric measurement for that, if you want.

The best way to deal with this problem is to find the bulk density. Weigh a dL of powder and see how much it weighs per mL by division. Be aware that any time one measures bulk density, things like packing density and so on enter into the picture. However, the measurement derived will be just fine for Black Powder. Heck, 55-60 percent of a load remains unburned anyway, so why agonize over too many decimal places? That derived number is close enough to use for both metric and Avoirdupois conversions. Powder is not measured by either the Troy system or the apothecaries' system, even though there are some units with the same names (but not necessarily the same weight)and the grains used are the same. Its measurement by the metric system is relatively new; however, it was measured by some other country-specific measurements, historically.

Once you have derived the number, it probably will be consistent for that lot of powder, but the next time you buy powder, it will probably change a little. I use 0.85 as a practical approximation, and it only really makes a big difference if one is shooting a large volume of powder. However, even 17 grains of FFFg, a load for my 1862 Police, measured by weight, approximates 20 grains by volume, which is still just fine.

As far as always using weight is concerned: It seems not to make any real difference, as long as one is consistent as to his method. Maybe it is the large unburned fraction, maybe the powder gods simply have decreed it that way; I don't know. Be aware, however, that even if you weigh charges you will probably need to make some adjustments when changing brands or lots of powder.

May 10, 2007, 06:48 PM
Bad Flinch,

I never knew there were differences in the metric system.
Since I went to school a cc and ml were always the same (I'm 45).
I'll dive into history, always interesting. Thanks.

If I understand correctly measuring by volume can still be done, as one actually still does today with the powder measure, after weighing the powder.
Or by using a simple scoop.

It is just that several people have told me you should never weigh black powder but use 'volume grains', and I never understood why.
As I understand now there is no reason what makes it really necessary.

Thanks again for the explanation, appriciated!:)

May 10, 2007, 08:07 PM
Pyrodex should be measured by volume generally. If you measure it by volume equal to the amount of blackpowder the gun needs, you get the right amount. Go by weight and you can get too much for the gun...

May 10, 2007, 08:31 PM
FYI...I have a couple of spare TC patent breech plugs and know that they each hold 18grns Goex 3F.

There is no need to worry about airspace with such a tiny charge of 10, 15, 20 grns of powder under a ball sitting on top of a Patent Breech.

People who occasionally "dry-ball" routinely put in that much powder to "bloop" out the dry ball...done it several times over the years myself...there's not enough powder to generate enough pressure for it to be an issue with regards to an air space / bore obstruction that I assume is being referred to here.

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