Quick Statics Question

[clh7837]

I have a vehivle that I want to weigh.

I pick the front end off of the ground and get a weight from the hoist.
I do the same with the back end of the vehicle.
I measure from centerline of each axle to the pick up point and the wheel base.

Does it make sense to sum forces and moments to get the weight that is on each axle? Then adding the weight on the two axles will giv me my weight?

Thanks, this is just kind of messing with me.

 

[rb1957]

if the CG is w behind the front axle, and L is the distance between the front and rear axles, and Pf is the load on the front and Pr the load on the rear lift and W the weight of the car; then ...
Pf*L = W*(L-w) and
Pr*L = W*w.
so w = Pr*L/W
and Pf*L = W*(L-Pr*L/W)
Pf*L = WL-Pr*L
gosh ... Pf+Pr = W (possibly should've expected that, but I didn't)

another day in paradise, or is paradise one day closer ?

 

[clh7837]

That comes out to just adding the measured weight from each end. Which I know is not correct. Maybe you didnt quite understand the problem.

I am picking up the front, 43 inches away from the front axle. Back end stays on the ground. This measures 1637 lbs
Then I pick up the rear, 18 inches from the rear axle. Front stays on the ground. This measures 2728 lbs
The wheelbase of the axles is 114 inches (centerline to centerline). The length from pick up point to pick up point is 114 + 18 + 43 = 175

What is the total weight of the car?

 

[hydtools]

You simply sum both measured weights to find total weight. If you want to calculate weight on the axles, then you have to do the moment math.

Ted

 

[rb1957]

"That comes out to just adding the measured weight from each end. Which I know is not correct." Why do you know it's incorrect ? There's a bit of judgement in getting the loads (Pf and Pr) right; but the math looks reasonable.

but now you're explained the problem better, L in the two equations is not the same. I read you were lifting at the axles, but now you're not. So CG w behind the front axle; then
Pf*(43+114) = W*(114-w) and
Pr*(18+114) = W*w
so w = Pr*(18+114)/W
and Pf*(43+114) = W*(114-(Pr*(18+114)/W) )
Pf*(43+114)/114 + Pr*(18+114)/114 = W
1637*1.3772+2728*1.1579 = 5413 lbs

another day in paradise, or is paradise one day closer ?

 

[gruntguru]

Alternative method.
Moments about the rear axle.
Pf x 114 = 1637 x (114 + 43) so Pf = 2254.5

Moments about the front axle.
Pr x 114 = 2728 x (114 + 18) so Pr = 3158.7

Total weight = Pf + Pr = 2254.5 + 3158.7 = 5413.2


je suis charlie

 

[clh7837]

Awesome, thanks for both of the answers. Easy way to weigh something without having large scales.

 

[Dougt115]

Indeterminate as far as I can figure. I can tell you that it weighs more than the 5413 lbs.

The method described will only give you an under approximation of the vehicle weight. As it assumes that the weight of the trunk and front end do not matter in each equation. (The moment of the car extending 43 inches out from the axle decreases the measured weight when lifting from the other end.) This method gives the best results when the axles are at the very ends of the vehicle or are symmetrical.

Alternately you can consider the fact that the center of gravity is different when each end is lifted because the pivot points are not symmetric. These methods assumed that the CGs were the same.

Consider this: If the axles were extremely close together the force measured at each end would be almost zero. And to the extreme case using the method provided if the car balanced on one axle at the CG the sum of the lifts would be zero and your car would be weightless.

This can be seen in the case where the CG is over the axle so that w=0 and from the equations used above
w=Pf*(114+18)/W Therefore Pf must equal zero and the weight of the car, W, can be number.

 

[rb1957]

don't get it ...
the CG of the car accounts for the total weight of the car. sure you could divide it up into portions ... ahead of the fwd axle, between the axles and behind the rear axle. sure weight on the front bumper pulls the CG fwd; so what. for calculating external reactions, if I know the CG of a bunch of weights, then applying the total weight at the cg is equivalent to applying the bunch of weights; obviously if you're interested in the internal stresses you need to consider the bunch of weights. if I have a SS beam with a triangular distributed load, I can calculate the reactions from the equivalent load at the CG; and it doesn't matter if the beam has overhangs.

another day in paradise, or is paradise one day closer ?

 

[Dougt115]

You are right until you try to find the loads my lifting the ends and you don't know the location or value of the CG. Then the uneven overhangs matter. The over hangs are giving you false readings in a large part because they are not symmetric.

Try changing the values of the over hangs of the center section, 114, to extremes and you will see the equations fail reality.

w = Pr*(114+18)/W and w=114-Pf*(114+43)/W

if 114 goes to zero wouldn't the axles line up with the CG? Then the forces needed at each end to lift would be zero and the sum of the forces would be zero so your car would weigh zero???

And if it goes large
Pr*(large number+18)/W = (large number)-Pf*(Large number + 43)/W

With (large number) >>>18 or 43

Becomes Pr * Large number/W = Large number -Pf(large number)/W
Large number * (Pr+Pf)/W = Large number
(Pr+Pf)/W = 1
(Pr+Pf)= W
Funny that this is only true when the distance between the axles makes the front and rear over hangs insignificant.

 

[rb1957]

if 114 goes to zero, then there's no wheelbase.

if 114 goes large then the effect of the overhangs reduces, so what.

my initial post assumed lifting at the axles (no overhang) and got Pf+Pr = W (what you get from your equations if 18 and 43 go to zero); as you've derived too.

solve a SS beam, without overhangs, for a triangular load and an equivalent point load (same total load at the triangle's CG); you should get the same results. now a SS beam with overhangs and the same two loadings ... again you should get the same results (though obviously different from the preceding case).


another day in paradise, or is paradise one day closer ?

 

[marty007]

Draw a freebody diagram... 5413.2 lbs is correct.

 

[desertfox]

Hi

I agree with Dougt115 the method will only give an estimate.
If you lift one end and take the load reading, then as he stated that load will be reduced by the over hang at the opposite axle, however we base our calculation on the reading we have just taken and assume that that reading applies when we put the end back onto the floor, we then use it to calculate the vertical reaction at the first axle, this is not true because the overhang now adds some weight to the vertical reaction that were trying to calculate whereas when we lifted the end of the truck that overhang weight assisted.
I would post some diagrams but I'm not in a position to do that till the end of the week.

 

[marty007]

Dougt115 and desertfox, the exact location of the COG does not matter, please see my proof attached. The weight is 5413.2 lbs.

Cheers,

 

[desertfox]

Hi marty007

Thanks for your efforts, I got the same result myself and was pondering over it for a couple of days before posting, the problem isn't the centre of gravity it's the assumption that after you measure the load you assume the measured value stays the same after the truck is put down however that load would slightly increase at the axle once you put the truck down because the overhang mass or weight over the axle your pivoting about no longer assists you.
at first I agreed that it was 5413lbs but when I thought it through I agree with Doug but I still think the method gives a reasonable estimate.

 

[rb1957]

I think the inaccuracy in the method is the value of the lift loads ... it'd be tricky to get the load just as the wheels clear. And sure the analysis is a bit theoretical (as opposed to heretical) ... ground contact points will change slightly with the lift, etc ... it depends on how accurate the result needs to be.

another day in paradise, or is paradise one day closer ?

 

[Dougt115]

This has been fun. Though my statements were not wrong they did over analyze a simple problem.

The car weighs 5413.2 lbs.

 

[clh7837]

It is actually a tractor haha. Its 3000 HP and we run in a 6500lb weight class. We changed rearends and were curious to see how much weight we lost. We will be weighing on scales April 25th, so I will report back on how close the actual is to calcualted. (if the scales are right!)

Thanks!

 

[gruntguru]

The main source of error in the method is translation of the CG due to its vertical position. The error disappears if either:
a) The CG, the lifting point and the rear axle centreline are co-linear in side view. or
b) The lifting process does not rotate the vehicle significantly

je suis charlie

 

[IFRs]

Old high-school class: carefully measure the pressure in the tires. Lift the rear tires off the ground, put carbon paper under the wheels, lower the car back to the ground, pick up the car again and remove the carbon paper. Repeat for the front wheels. Measure the area that the tires made on the paper. Multiply by the air pressure.

On the other hand, since you are only trying to get the difference for the new rear end, lift the rear off the ground before and after the rear end is replaced. Lift at the wheels, not on the bumper. It does not matter what overhang there is or where the CG is. The difference in weights will be the rear end.