On Thu, 24 Jun 2010 21:54:18 -0700, Winston
wrote:

On 6/24/2010 5:21 PM, Ned Simmons wrote:
On Thu, 24 Jun 2010 11:01:48 -0700,
wrote:

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:

(...)

Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

In this case, yes.


I thought the seams represented the weakest link.

I don't know that for a fact, but let's assume they are. Two inches of
weak seam will resist more force than one inch.

The force per inch is the same in both cases, yes?

I don't grok.


Perhaps we have different cases in mind. My point is only that the if
the pressure and area remain constant while the seam length increases,
the force per unit length of seam will decrease. Clearly this is
meaningless if we're limited to cylindrical drums, because there's a
fixed relationship between head diameter and circumference.

force/length of seam = pressure x area / seam length
or
lb/in = lb/in^2 x in^2 / in


--
Ned Simmons