It's stupid, I know, but...

[StrangeRanger]

That's why you have to use mass instead of weight. Mass is an inherent property. The mass of an object is the same anywhere in the universe, weight is the effect of a gravitational pull acting on that mass. The relation is W = M*g. G on earth is 32.17 ft/sec ^2; on the moon it is approx. 1/6 of that or 5.4 ft/sec^2.

One of the advantages of the metric system is that it's basic unit the kilogram is a mass unit. A 1 Kg mass is the same on the Earth or the Moon. It weighs 2.2046 lbs. (or 9.81 Newtons to use the metric weight unit) on Earth and about 0.34 lbs. (1.51 N) on the moon. All dynamics equations are concerned with mass not weight so things are a lot easier in metrics. (Admittedly if you work mostly in statics and strength of materials, Imperial units are much more convenient.)

 

[StrangeRanger]

SM, I missed your last post, concurrent posting. You're right I simply used the "earth" equations and changed g. I needed to first calculate the earth masses in slugs (really, don't ask!) and then plug those into the lunar equations. Look for an edit later tonight.

The second equation is KE = PE

The kinetic energy gained translates into potential energy available. 100 ft-lbs. of KE wil be enough to lift 100 lbs. 1 Ft.

 

[broncr]

So a turbo/vacuum setup would work better on the moon? ( )

 

[ford6man27]

Getting back to hardcore inline tech, how far would a 300 six move [on the moon] if you clamped a 357 magnum to the intake and fired it? That is, with all assecories including flywheel and clutch,and with the gun pointed forward in line with the engine. Andy

 

[StrangeRanger]

OK the edit is posted. Very little difference overall. The errors largely cancelled one another out. The powder velocity is the mean velocity, approx 1/2 the velocity of the front which is where the sabot would be riding, i'll stick with 2500 until proven wrong - which may not be a long time at the rate I've been going lately..

 

[SuperMag]

H = .0044 ft or about .05"

You're still going to need that .378 Weatherby......

SR, I went through the same exercise using the recoil formula found on the Hodgdon powder site FAQ...

Recoil Energy in ft. lbs. = (Bw Mv + 4700 Pw)^2 / (64.348 Gw) where:
Gw= the weight of the gun in pounds
Bw = weight of the ejecta (shot and wad) in pounds
Pw = powder weight in pounds
Mv = muzzle velocity in feet per second

I'm assuming that the expansion velocity of gunpowder is the 4700 figure. Anyway, I substituted 335lbs in for the weight of the gun, and plugged the resulting figure into the KE=PE formula (with the lunar acceleration, of course), and the Height I get is 0.130"

I wish I could remember the context of the original scenario (I was in junior high when I read it). But based on the exercise here, it would seem that I read it in an Isaac Asimov novel...

 

[StrangeRanger]

Asimov was too good a scientist; he would NEVER have made a mistake like that.

 

[Anonymous]

Getting back to the original question, you don't really need to use any active mechanical device to scavenge the cyclinders. A properly designed exhaust system will do it for you. David Vizard wrote a great article about how to do it yourself some years ago, and someone scanned into this geocities page.