Hey Dave,
Exactly 5 years ago to the day we were discussing this same subject at
http://listerengine.com/smf/index.php?topic=1431.msg19154#msg19154Was reading through it and enjoying the exchange between Bob and GuyFawkes, that old blowhard. I really miss him. Much as I hate to admit it, I learned a lot from his rants.
Anyway, was thinking about balancing again. And I read that you wrote:
Re: engine mounting v 4.0
« Reply #104 on: December 29, 2006, 03:09:39 am »
Reply with quote
Quinn,
Yes it is nice to see some other numbers. I should have posted my Vidhata #'s also but didn't want to confuse the issue.
Check your addition... I added yours several times and could not get the same numbers. ?!!!?
VIDHATA
Small end of Rod 1216 grams = 2.680#
Piston assembly 4025 grams = 8.873#
Big end of Rod 2742 grams = 6.045#
-----------------------------------------------------------
Total 7983 grams = 17.599#
Bobweight @ 65% 6149 grams = 13.556#
POWER SOLUTIONS Kit
Small end of Rod 1147 grams= 2.529#
Piston assembly 4024 grams= 8.871#
Big end of Rod 2584 grams= 5.697#
-------------------------------------------------------------
Total 7755 grams= 17.096#
Bobweight @ 65% 5945 grams= 13.106#
Quinns
Small end of Rod 1194 grams= 2.632#
Piston assembly 3840 grams= 8.465#
Big end of Rod 2819 grams= 6.214#
--------------------------------------------------------------
Total 7853 grams= 17.312# HuhHuh I think I added it up 5 times same answer??!!
Bobweight @ 65% 6091 grams= 13.428#
avg assembly wt. 7864 grams= 17.335#
avg bobweight @ 65% 6062 grams= 13.363#
I am betting the differences in the assemblys will only be a small % in the balance factor another words my flywheels would run well on your engine. I do need to put my P/S wheels on the Vidhata to confirm that!After I figured out you were measuring 65% of the reciprocating portion, not the whole enchilada we were in agreement with your numbers. You should be able to calculate this easily. It's a simple torque problem. The torque imparted to the crankshaft by the bobweight at half the stroke length (5.5" / 2 = 2.75" radius) from crankshaft center must exactly balance the torque imparted in the other direction by the weight of the counterweights (2 of them) on the flywheels. I measured that radius at 10" from the crankshaft. That might be a little off, and the engine is 1100 miles away now and I won't be there until the middle of January so I can't measure it right now.
So the torque diagram would look like this:
10" 2.75"
+----------------------------------------------------O-------------------+
1691g ^ 6149g
So that means that at a radius of 10" from the center of the crankshaft, 1691g will balance out the 6149g bobweight you placed on the crank pin which is centered 2.75" from the center of the crankshaft.
That means that each flywheel needs 1691g/2 = 846g of weight, or 28.9 ozs of weight. That's a lot less than both Jack and I measured as the weight of the cast in weights on our flywheels. As I recall Jack measured his at 47 and 48 ozs, while mine were 35 and 54 ozs. or about 45 ozs average.
Maybe the 10" radius is off, but if it is, it's not off by much. My engine's at the wife's in Washington, so can't nip outside and check that radius at the moment.
I'm sure there's some bright individual with nothing to do on New Year's eve-eve but tell me where I'm off. Any thoughts?
Quinn