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NitroXAdministrator
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The Greenhill Formula for calculating barrel twist rate
      #368490 - 21/08/22 05:20 PM

Extracted from another thread.

Here's the Greenhill formula for those who might want it; but please take note of the quote from Daryl S's post above:

"Many fail to remember or realize it is showing the slowest twist that will work, not the optimum rate of twist"

The Greenhill formula

T= (150 x D)/R

Where:
T- is the twist required (number of inches per one revolution)
D- is the diameter of the bullet (in inches)
R- is the ratio of bullet length to bullet diameter (length divided by diameter)

Conversely, to find out what length bullet will be stabilized in a
given twist, use:

L= (150 x D x D)/T

Where:
L- is bullet length

The number 150 is a constant used by Greenhill, and works well at velocities in the vicinity of 1500 feet-per-second or greater. At 2800 feet-per-second the constant can be changed to 180 with good results.

Note that it is the bullet length that is important, not the bullet weight. Greenhill works well with all lead and lead-alloys commonly used for bullets.
One must also understand that Greenhill is derived from experience as well as from the calculated laws of physics. As such, it is a highly simplified, albeit useful equation. Greenhill was worked-out many years ago and is quoted in the
“British Textbook of Small Arms” (1929)

Hope this helps

Best
ALEX

From this thread:
http://forums.nitroexpress.com/showflat.php?Cat=0&Number=180654&page=0&fpart=3&vc=1

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NitroXAdministrator
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Reged: 25/12/02
Posts: 39877
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Re: The Greenhill Formula for calculating barrel twist rate [Re: NitroX]
      #368491 - 21/08/22 05:23 PM

Re the Greenhill Formula, my lack of knowledge of physics might show but I would have thought bullet weight might have been relevant as well in a formula?

But maybe not.

--------------------
John aka NitroX

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Govt get out of our lives NOW!
"I love the smell of cordite in the morning."
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NitroXAdministrator
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Reged: 25/12/02
Posts: 39877
Loc: Barossa Valley, South Australi...
Re: The Greenhill Formula for calculating barrel twist rate [Re: NitroX]
      #368492 - 21/08/22 05:37 PM

Let's try it.

A Woodleigh .333 300 gr SP. 1.459".


T= (150 x D)/R

Where:
T- is the twist required (number of inches per one revolution)
D- is the diameter of the bullet (in inches)
R- is the ratio of bullet length to bullet diameter (length divided by diameter)

(150x.333)/(1.459/.333)=11.4"

If it is resized to .330 = 49.5/4.4212 = 11.2"
assuming no additional length.

A .330 250 gr resized to .323" bore:

1.302" 250gr SP .330 = 48.45/4.0310 = 11.24"
1.419" 250gr FMJ .330 = 48.45/4.3932 = 11.03"
i.e. for resized to a .323 bore.

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"I love the smell of cordite in the morning."
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LRF
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Re: The Greenhill Formula for calculating barrel twist rate [Re: NitroX]
      #368518 - 22/08/22 06:39 AM

Quote:

Re the Greenhill Formula, my lack of knowledge of physics might show but I would have thought bullet weight might have been relevant as well in a formula?

But maybe not.


Greenhill is a 25,000 foot high estimate. When getting down to earth bullet weight, bullet shape and speed all play into a fine tune twist.

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