"Would a "real" 28mm scaled Martini-Henry bullet kill you?" Topic

The OFM got me thinking in another post. I should be working…but I'm "haunted" by this question.

Say someone created an actual, working Martini-Henry rifle in 28mm scale (size) with compatible bullets, etc. and that you could actually fire it.

What kind of damage would a projectile that size -traveling at the Martini-Henry muzzle velocity (whatever that is) do to you if anything?

Maybe a completely dumb question – but inquiring minds want to know!

;-)

You'll put your eye out with that thing!

Not sure but I will try it later and LYK

Having dreams about all those Brits and Zulus coming alive?

Hi

1300fps, but the round would be smaller than a grain of sand and have a mass of the inverse cube. Not enough kenetic energy to notice.

R

I would think that, given that the bullets would be about as thick as a pin and have a microscopic amount of gunpowder in them, the damage would be not a lot! But there are other far more educated in physics, engineering and firearms on here than I am!

My father in law fired a real one and told me it blew a truly massive hole in the target (Not a Zulu!!!!!!).

Not that silly of a question really, whenever I see a giant peppered by arrows in a movie I wonder how much of an effect it would have.

I suppose if a mini-martini-henry bullet hit you in the eye, you'd notice it.

This sounds like a job for Mythbusters!

Mass x velocity = impact, right? Calculate the mass of the scaled bullet (no idea) x what you believe the velocity might be (you'd have to calculate scale range) and you'd get your impact.

Believe that I read somewhere years ago that 50 foot pounds will knock a man down. Mass of the bullet and decreasing muzzle velocity can be calculated to show impact at various ranges.

Dan

Mass x Velocity Squared, my good man…..

it would not have enough energy to penetrate skin
BB from a basic air rife cant even do it and they have
a lot more energy then a 28mm m16 would or Martini-Henery

grain of sand moving at about 14 feet per sec.

This has Mythbusters written all over it!

I think you have been listening to too much Oingo Boingo.

I couldn't even imagine how you would crush the ingredients to make the gunpowder that small.

Now if they were humans shrunk down to the size of 28mm.

Stephen King did his model soldiers attacking the hitman in his apartment. The choppers and bazooka did all the damage.

I think a 28mm scale cannon wouldn't do much damage either, at worst it would hit like a BB.

Don't forget the tactical Nuke in the box…

The bullet would bounce off the hairs on my manly chest.

Seriously, if the muzzle velocity is down by 1/72 (just to grab a number…), the MASS of the bullet will be reduced by the CUBE of 1/72.
I do not think I would even feel a chest hair being hit.

Would a 28mm Martini-Henry have enough muzzle velocity to overcome the drag of wind resistance between the muzzle and you?

Now a 28mm scale sword bayonet, that could hurt. Might not do lasting harm, but it would case some pain.

did someone say Martini's… ? ..8^D

Thanks, Top Gun Ace, knew there was a reason I majored in history (smile).

Dan

Ok if I've got the math right the bullet would be 0.039 inches in diamiter

Well I'll add to this nonsense with some REAL number crunching! (Never get an engineer stirred up on scale stuff!)

Using the Wikipedia entry for the Martini-Henry as a basis and assuming we're talking about the original Snider cartridge, the round contained 85 grains (5.5 grams) of gunpowder, the bullet weighed 480 grains (31 grams), and the muzzle velocity was 1350 ft/sec (410 m.sec).

For ease of calculation, I'm assuming 28 mm scale is equivalent to 1/50 scale (1" on the model = 50 inches real life and I will let you all argue over what scale ratio is really meant by 28 mm). The Snider cartridge had a .577 caliber round (0.577 inch diameter) giving a 1/50 scale size of .01154 caliber. Weight is dependent upon mass and volume. Volume requires three dimensions and, if all dimensions are held in a constant ratio when applying a scaling factor, the reduced volume is determined by the cube of the scaling factor. The cube of 1/50 (0.02) is 0.000008 (or 8.0E-06). So all weights must be multiplied by the cube to obtain the reduced scale weight (I see those eyes glazing!).

Your 28 mm Snider cartridge would contain 44 milligrams (or 0.00068 grains) of gunpowder (compared to the full size quantity of 5.5 grams or 85 grains) and the bullet would weigh 248 milligrams. I think I can safely say that 44 milligrams of gunpowder is not going to propel 248 milligrams to a speed of 1350 ft/sec (or even a 1/50 scale of 27 ft/sec).

As far as kinetic energy is concerned (MV2), if the mass is reduced by the cube and the velocity is reduced by 1/50 and then squared (0.0004), the KE would be reduced by 3.2E-9.

Reality is such a bummer compared to make believe.

gamertom, your masses are off by three orders of magnitude. The cartridge would weigh 248 micrograms.

In my (not so humble) opinion, at 1:50 scale or so, the friction inside the barrel would stop the bullet ever getting out of the barrel. But indeed that little lead (or other metal/plastic) bayonet would really smart.

pigbear, you are quite right, 2.48E-04 grams is 248 micrograms and the gunpowder weight would be 4.4E-05 grams (or 44 micrograms). Got my "micro" and "milli" terms backwards in typing. Should have stuck with scientific notation.

And this is exactly why I LOVE TMP!

Great to learn something new today!

So what would happen to a 50 foot giant hit by a real size Martini bullet?

Hi

You mean a 50 giant as in magic fantasy, or a real giant unable to even move because of the weight causing his legs to collapse?

R

Dont start waving physics in front of me now. I dont remember the formula off the top of my head about mass and volume as you increase volume. I know King Kong would be unable to support his own weight. So let me restate the question.

What would a Martinin bullet do to KIng Kong as he lays on the ground, crushed by his own weight.

OOOOH mathematically geekery at it's finest. Awesome!

gamertom and pigbear you rock for actually figuring it out.

Actually, I was reading a site using physics to show that movie monsters couldn't exist and it came to the conclusion that the upper size limit for a gorilla would be about 25', ie King Kong size. He wouldn't be as agile as the Peter Jackson version but he wouldn't collapse under his own weight either. Most other movie monsters didn't fare so well.

Also, unless the winter of 1930 was *seriously* Arctic, the ice on the pond in Central park wouldn't support Kong's weight, despite the scene in the recent remake. So instead of having a touchingly clumsy pas-de-deux with Naomi Watts, Kong would have ended up armpit-deep in nasty frigid New York pond water.

Far less cinematic.