Forum Discussion
Cummins12V98
Apr 23, 2020Explorer III
ford truck guy wrote:Cummins12V98 wrote:fj12ryder wrote:Cummins12V98 wrote:I think you mean 2.35%. .0235% of 6000 lbs. would be 1.41 lbs. :)
Hers is a simple example of how to determine how much weight you will add to your front axle depending on pin placement VS center of rear axle.
My truck has a 170" wheelbase, if i move my hitch 4" forward that will mean I will transfer .0235% of my 6k pin weight or 141#. This does jive with my loaded and unloaded front axle weights.
I'm not really sure it would work that way, 4 inches is 2.35% of your wheelbase, but not sure that weight transfer would be that linear. But I dunno.
Picky, Picky! :B
4 divided by 170 is .0235%, or .0235 x 170 is 4. Take 6000x.0235 = 141.
Sure does work and I do believe my math is correct.
2.35% of 6000# is in fact 141... converting 2.35% to decimal form takes it to .0235...
Your both right
It can't be ETHYL!!!
Thanks, I used my "BASIC" calculator.
About Fifth Wheel Group
19,006 PostsLatest Activity: Feb 18, 2025