For you engineer types

[Rifleman]

Ok guys maybe one of you engineer types can answer this for me (hey bluedot 37.5 )
If I cut off 4 inches from a blued 10mlII barrel from the muzzle end, how much would the cut off piece weigh?
Thanks Rifleman

 

[Joe Name]

Twice as much as 2 inches.

 

[flatland hunter]

Bad, Bad man :lol:

 

[Blue-Dot-37.5]

Man Dwight, now you're making me think! How 'bout I let you figure it out?

Area of a circle = Pie (I know you'd like this formula! ) x radius squared ( in inches). times 4. Do that for the outside of the barrel. Now do that for the inside of the barrel and subtract that from the OD number. Now multiply that by .2833. That's how much it weighs. (in pounds)

Or just throw it on a scale and weigh it! 8)

Blue-Dot-37.5

 

[Anonymous]

4140C weighs about .283 lb. / cu. in., which I think BlueDot more or less said. So, the rest is just fun math excercises. 8)

 

[Rifleman]

Area of a circle = Pie (I know you'd like this formula! ) x radius squared ( in inches). times 4. Do that for the outside of the barrel. Now do that for the inside of the barrel and subtract that from the OD number. Now multiply that by .2833. That's how much it weighs. (in pounds

WHUT? :huh?: What about the taper?

Or just throw it on a scale and weigh it!

I was kinda wanting to know the answer before I made any cuts :shock:


Nice try Joe Name but I think these guys might be closer.

 

[Joe Name]

Sorry R-Man I just could not resist.

 

[Blue-Dot-37.5]

WHUT? :huh?: What about the taper? quote]

Sheesh!! :mrgreen:

Gimme some dimensions and I'll get back to ya after dinner..... :lol:

Blue-Dot-37.5

RW:

You are right, I use a rough guide of .2833#/cu. in. for steel and .097#/cu. in. for aluminum. Some alloys will vary, but not enough to matter (much).

BD

 

[FWF]

Taper?!?

Aren't we a little anal today? :wink:

Take the OD at the midpoint; that should get you close enough.

 

[Underclocked]

The inherent error in this method due to weight reduction in the tremendous taper should be offset by the weight of the lands. :wink: :lol:

 

[savagebrother]

uc,
what the h--l you doing over her????
antagonizin rw?? lol!!!!!!!
sb

 

[Underclocked]

Nobody said I couldn't. :lol:

 

[Rifleman]

 

[Rifleman]

 

[Hunt 4 Bucks]

Ok, I will bite on this.

By my measurements, the barrel dia. is appox. 13/16?

The bore is appox 1/2? (average considering lands are narrower)

3.14 (0.405) squared = 0.50 inches squared

3.14 (.25) squared = 0.19 inches squared

0.50 ? 0.19 = 0.31 inches squared

0.31 X 4 inches = 1.24 Cu. In.

1.24 X 0.283 = 0.35 pounds or 5.6 ounces (not perfect, but close)

Now this does not consider the weight of the front sight if you have one.

 

[Anonymous]

The inherent error in this method due to weight reduction in the tremendous taper should be offset by the weight of the lands.

And seas, particularly on mono-polar non-alkaline ionic alpha water, or whatever that stuff is. :shock:

 

[Underclocked]

Precisely.