Surface area on back end vs velocity

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When you run a hand puwered ports power if you have no load you can pump fast , but as load increases you cannot pump as fast because load also increases on source of power be it hand or motor
 
Have 5 each ,flat base ,& boattail loaded I identical , everything weighed the same including finished rounds will set up and test soon. Thank you for all thoughts so far , but right now still up for grabs
 
Assuming you have a chronograph. You will need to document each type of bullet's bearing surface area to account for frictional differences. Also assuming all bullets are from the same manufacturer to eliminate variations in jacket materials. You won't be able to document the jacket thickness variation unless you section and measure. All will affect the resistance of the bullet to deform to the lands and affect peak pressure/muzzle velocity.

They may be small enough to be lost in the velocity spread, but you'll need it just in case.
 
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I get what you are saying, it just isn't correct. True you get more surface area and thus more force, but the mistake you are making is thinking that all of that force helps you in the direction you want it to. The pressure acts in a vector normal to the surface. You only get to count the component of that vector acting in the direction you want it to (ie inline with the ram axis). You've gotta use trig for that. Example would be suppose you had a 1 sq in. area at a 45 degree angle in an volume of 1psi liquid. Because it's at a 45-degree angle to your ram, the force on the area will be 1lb but only ~.70lb will be acting in the direction you want. Because you have more surface area, it will break even however. Your caluclation never takes into effect this vector component breakdown. You'd either need a formula to calculate it or a good knowledge of calculus to do so depending on the shape you go with.
 
What everybody is trying to say (or has said) is that you're talking about increasing exposed surface area without increasing cross sectional area. Surface area doesn't matter, cross sectional area does.
 
The boat tails I use seem to give slightly higher pressure than flat based of same weight but I don't know whether it's because of the slightly reduced case capacity or because of the heavy copper bases. Maybe it's a little of both.
 
Everything else being equal (weight, ogive) a boattail bullet is slightly longer than a nonboattail.
 
From eldon519
I get what you are saying, it just isn't correct. True you get more surface area and thus more force, but the mistake you are making is thinking that all of that force helps you in the direction you want it to. The pressure acts in a vector normal to the surface. You only get to count the component of that vector acting in the direction you want it to (ie inline with the ram axis). You've gotta use trig for that. Example would be suppose you had a 1 sq in. area at a 45 degree angle in an volume of 1psi liquid. Because it's at a 45-degree angle to your ram, the force on the area will be 1lb but only ~.70lb will be acting in the direction you want. Because you have more surface area, it will break even however. Your calculation never takes into effect this vector component breakdown. You'd either need a formula to calculate it or a good knowledge of calculus to do so depending on the shape you go with

Bingo, I agree 100%.

Yes boat tails develop more velocity down the bore but my belief is that it's due more because of the decrease in usable case volume rather than the increase in surface area.
 
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