On 15 May 2005 10:00:49 -0700, wrote:

Hi everyone -
I have read through quite a few of the strings on inertia and moments
and am a confused lay-person. I believe the entries were repeatedly
saying that it is not as easy to calculate the inertia/mass per length
of moment, especially adding angle .... but forgive me for asking
again.

After a back surgery, I am building avery simple shop crane from square
tube stock. It is modeled mostly after engine hoists, but without the
ram. It is toned down strength-wise because it simply has to hold
250lbs about 45" out on an arm (rather than the 1 and 2 tons an engine
hoist holds) - and I am probably overbuilding.

There are no hinges, tube stock sleeved 20" into larger tube stock will
allow me to make basic angle/height adjustments with a pulley doing the
rest of the up and down work.

Not sure what you mean here. You say "without the ram". Do you mean
the arm is supported only at the end, or that you have replaced the
ram with a sleeved tube, that can be adjusted for gross height? (I
hope the second). The reason I ask is the the sleeved tube _could_
simply slide up and down in the upright portion and you say "no
hinges".

I am also not quite happy about how you are mounting ghe whole thing.
Forces get larger as you move away from the load, in this case. The
base of the upright, or whatever, could be taking more strain than the
arm itself.

Can you do some "ascii art" to show the layout? Or post some rough
drawings to either the metalwork binaries group, or to a website?

I have seen calculations like I=m(r)squared, but am unsure what length
the "r" is. Have also seeen I=(m)(1)squared/3. Anyway, for a simple
structure as this .... ok ... here it goes .... is there a simple way
to calculate aprox. inertia/mass at the end of the arm to ensure the
tube stock can more than handle it (because of fluctuations from
bounces and jostles when moving about) - thanks!