Weighing both ends of a connecting rod

nyholku

Third time lucky, what ever happened to my two previous attempts to post this...

anyway, I found an ancient approximate formula to partially balance an IC engine
that requires weighing both ends of the connecting rod like this:

http://www.modelenginenews.org/etw/etw_bal/images/bal_p4.jpg

It is too long since I took Statisc 101 so I have hard time figuring out how to
do this weighing in Onshape:


Here is a dummy connecting rod I created for the exercise:


https://cad.onshape.com/documents/fba55f39fa348ca0d0087d9c/w/8e676d47bfff6e16d117f23b/e/4b6e81178b362f44874f1448

NeilCooke

Are you trying to find the centre of mass? Apply a material to the part (right click) then select the part in the parts list and then press the scales icon in the bottom right corner. 

nyholku

Thanks for taking interest.

The thing is, I don't rightly know! 

The method is described here if you are interested:

http://www.modelenginenews.org/etw/etw_bal/p2.html

In that method you are called to weight the connecting rod as in the first picture I linked to.

I know I can get the center of mass from Onshape easily. But how would I use that to
'measure' how the connecting rod would be weighted by two scales if balanced as 
in the picture?

My intuition says that CM somehow comes into it but it has been too long since I did this sort of thing and never in anger.

If I find the CM and cut the piece vertically at that point would the weight of the separate pieces be what I'm looking for?

NeilCooke

I don’t know. It is supposed to be balanced so the weight is the same? Or is there a ratio between the big and small end? I will read. 

ChuckKey

Use moments. If the mass is M, the bearing centres are L apart, and the centre of mass is X from the big end, then the weight at the little end is MgX/L and at the big end Mg(L-X)/L.

nyholku

Thanks ChuckKey, of course, a little refresher in statics would not hurt me obvious in hindsight!