SHINTON
Well-Known Member
- Joined
- Mar 17, 2005
- Location
- Triad area of NC
Ok...this might be a math major/engineer type question. Thus the reason it is in tech area...
Assuming for a moment here, single seat buggy, front engine/tcase/tranny setup.
I am going to guess weight as 2000lbs, maybe less.
I am also going to guess 65% of weight on front? (I know they have to be careful with this) at most 60%
Ok...so now...assuming you are going down an incline of 25 degrees. That shifts the center of gravity fwd...and thus places MORE pressure/weight on the front tires...
Any way to estimate/figure out the amount? Say it was 95" wheelbase?
My limited math guesses here is this:
At 90 degrees (yeah i know)...100% of the weight would be on the front tires...in fact I am going to assume that at 75 degrees for our purposes.
So at 0 it is 65%, 75 it is at 100%
so at 25 it would be at 35/3 = 10.167 more, thus 76%?
Therefore at 25 degrees, 2000lbs, 1520 lbs on the front axles and 480lbs on the rear axle?
Ok...is that "close enough" for ballpark? I would imagine it is a much more complicated calc, with height of tires, wheelbase, etc, but wanted to get a guess.
NOW....what insane idea am I cooking up you ask?
Again...assume single seat buggy, as above...narrow axles (60" including tires wide), thus fitting inside the width of pickup bed rails.
You build a ramp, starting at edge of wheel wells, extending to top of pickup cab. Then goes "flat"/horizontal going fwd over the cab.
Now...you BACK up the buggy onto this ramp/rack system...the part that hangs over the cab would need to support 480lbs according to the calc above.
76% Weight of the buggy would be just behind rear axle. (about 20-24" maybe based on center of tire)
Anyway, now you know the insane idea...the goal would be keep the buggy as light as possible. The truck would be a 1 ton dually....which in theory can handle 2000lbs in the bed, correct?
Oh...just a picture of something similar:
(well heck, will be back later with pix)
Sam...scheming...Hinton
Assuming for a moment here, single seat buggy, front engine/tcase/tranny setup.
I am going to guess weight as 2000lbs, maybe less.
I am also going to guess 65% of weight on front? (I know they have to be careful with this) at most 60%
Ok...so now...assuming you are going down an incline of 25 degrees. That shifts the center of gravity fwd...and thus places MORE pressure/weight on the front tires...
Any way to estimate/figure out the amount? Say it was 95" wheelbase?
My limited math guesses here is this:
At 90 degrees (yeah i know)...100% of the weight would be on the front tires...in fact I am going to assume that at 75 degrees for our purposes.
So at 0 it is 65%, 75 it is at 100%
so at 25 it would be at 35/3 = 10.167 more, thus 76%?
Therefore at 25 degrees, 2000lbs, 1520 lbs on the front axles and 480lbs on the rear axle?
Ok...is that "close enough" for ballpark? I would imagine it is a much more complicated calc, with height of tires, wheelbase, etc, but wanted to get a guess.
NOW....what insane idea am I cooking up you ask?
Again...assume single seat buggy, as above...narrow axles (60" including tires wide), thus fitting inside the width of pickup bed rails.
You build a ramp, starting at edge of wheel wells, extending to top of pickup cab. Then goes "flat"/horizontal going fwd over the cab.
Now...you BACK up the buggy onto this ramp/rack system...the part that hangs over the cab would need to support 480lbs according to the calc above.
76% Weight of the buggy would be just behind rear axle. (about 20-24" maybe based on center of tire)
Anyway, now you know the insane idea...the goal would be keep the buggy as light as possible. The truck would be a 1 ton dually....which in theory can handle 2000lbs in the bed, correct?
Oh...just a picture of something similar:
(well heck, will be back later with pix)
Sam...scheming...Hinton