What's wrong with this simple method of weighing?

[excelrvguy]

When I was President of an RV manufacturing company, I weighed many trailers. I've always subscribed to a simple method of weighing our 5th wheels. I was often met with resistance and engaged in several debates about my method being wrong, but no one could ever explain why.

I never felt that weighing the truck & trailer, then weighing the truck empty, then weighing the trailer.... blah blah blah was an inefficient way to weigh, not to mention being a real hassle.

When I weigh a 5th wheel, I pull the unit across the scales until the rear truck tires are off the scales, but not the jacks of the 5th wheel. Then I let the jacks down on the unit, just BARELY taking the weight off the hitch. (Not enough to actually lift the hitch.. Like maybe 1/8" to where I could barely see daylight between the hitch plate and the pin box)

Then I get a weight stamp. This would be the total weight of the unit. I would raise the jacks back up and have them stamp the weight with only the axles on the scales (I haven't moved a thing).

Example = Total weight 14,500
weight on axles 11,600
leaves only the pin weight. 2,900

I'm not talking about weighing each axle or each wheel. To me, this method saves time and effort and hassle.

What say you?

[Vanished]

Thank you PUCampin - appreciate the response and clarification.. 🙂

[PUCampin]

Vanished wrote:
http://academic.brooklyn.cuny.edu/physics/sobel/Phys1/eqm.html

As D2 changes (say moving from pin to jacks) the sum of the moments still needs to be 0... F1 (weight on tires) does not change, only the F2 increases assuming the CG is in front of back wheels..

Using the link for nomenclature

If like you suggest F1 and D1 are fixed, and the moments about A are 0, then as D2 varies F2 inversely varies. In fact, if you make D2 the same as D1, then F2 would have to equal F1 for moments to remain zero. However this ignores one other thing we do know. The sum of ALL forces and moments must be zero. There are no forces in the X direction but there are in the Y direction. We also know the weight A does not change. We know F1 + F2 must always equal A in order for the forces to add to zero. So if F2 increases, F1 must decrease because the total weight of the trailer did not change and there are no other outside forces.


You have 2 unknowns, F1 and F2. You need 2 equations to solve for 2 unknowns.

For those interested,:) consider the following

Sum forces in Y

F1 + F2 - A = 0 or F1 + F2 = A

Sum Moments about a point = 0 (we will use A)

-F1D1 + F2D2 = 0 or F2D2 = F1D1


For simplicity sake lets say A = 15,000lbs. So F1 + F2 = 15,000, but you don't know exactly how much F1 or F2 is. That is where the moment equation comes in

F1D1 = F2D2 or F1 = F2(D2/D1) So F1 will always be a factor of F2 and depends on the ratio of D2/D1.

For exercise sake lets say
Axle to COG D1=5
COG to Pin D2 = 20

So F1(5) = F2(20), F1 = (20/5)F2, F1 = 4F2


Now go back to F1 + F2 = 15000

since F1=4F2, 4F2 + F2 = 15000, 5F2 = 15000, F2 = 3000 (pin)

and then F1 + F2 = 15000 so F1 = 15000-F2, F1 = 12000 (axles)


So what happens if we change D2?

If say the landing gear is 5 feet closer than the pin,
D2 = 15

Then D2/D1 will be (15/5) so F1 = 3F2

We know A stays the same, so

F1 + F2 =15000, 3F2 + F2 = 15,000, 4F2 = 15000, F2 = 3750 (landing gear)

F1 + 3750 = 15000, F1 = 11,250 (axle)


So by making D2 shorter, F2 increased and F1 decreased, but not by a significant amount. But if you keep moving the support closer, when D2 is the same as D1, then F1 will be the same as F2, each half of the trailer weight A

[hotpepperkid]

Most of us could care less what the pin weight is unless your towing a 40 footer with a Toyota. I weighted the steering axle, drive axle, each trailer axle all under the max rating for each axle and under the max weight rating of the tow vehicle that is all I care about

[Dave_H_M]

cool, you're good like my grandies say. :B

[Vanished]

http://academic.brooklyn.cuny.edu/physics/sobel/Phys1/eqm.html

As D2 changes (say moving from pin to jacks) the sum of the moments still needs to be 0... F1 (weight on tires) does not change, only the F2 increases assuming the CG is in front of back wheels..

[bpounds]

^1 on the above.

Or, putting it another way, think of the landing gear as a fulcrum. The pivot point on a lever. Whatever weight that is forward of the LG fulcrum is pulling that side of the lever down, which wants to lift the axle side of the lever.

And yes, it isn't much. Most recreational fivers are fairly light up front by design.

Would be a fun exercise on a slow day at the scale. Position the truck and trailer with the trailer axles on the scale, and the truck AND landing gear off the scale. Check the axle weight, and then raise the landing gear until the pin weight is off the truck. Then check the axles weight again, and you would see it has been reduced.

To exaggerate the idea, continue raising the landing gear and lift some of the truck itself. Actually, don't do that, because it is really bad for your landing gear.

[PUCampin]

Sitting on the landing gear absolutely removes weight from the axles vs the pin, but it isn't a lot of difference. If you know engineering it is just simple statics to compare them, solve the system of equations that sum the forces in the vertical direction and sum the moments about a select point.

Conceptually think of it like this. You know the center of gravity is forward of the axles because there is pin weight. So if you take the support at the pin and gradually move it back towards the axles, what happens? At some point the support is under the center of gravity and thus supporting and balancing the entire weight of the trailer, the axles are now not supporting anything and the trailer could tilt back and forth like a seesaw. So if you support it at the jacks, they will carry more load than if supported at the pin, so the axles support less. It won't be much different, on the order of a few hundred pounds, because the difference in distance from the center of gravity to the jacks vs the pin is small. The exact amount depends on the ratio of COG to pin divided by COG to Jacks

[Vanished]

We are talking two different issues/questions..
1) I agree total weight (if all on one scale) minus axle weight would give you pin weight. I misread and corrected previously
2) I still disagree that sitting on landing gear puts more/less weight on rear tires than they see when hitched up.

[Bowti]

rhagfo wrote:

Vanished wrote:

bpounds wrote:

SoCalDesertRider wrote:
...
Now, on the pin weight versus jack weight issue, if you were weighing just the jacks by themselves, no trailer axles on the scale, that would not be the same as the pin, because there is about an 8 foot length difference between the pin to axle measurement, versus the jack to axle measurement. The jack weight will be significantly higher than the pin weight.

Interesting thing about this statement (which is true), is that it means the weight on your tires is less when your trailer is sitting at camp, than when you are hitched up. The weight forward of the landing gear is lifting weight off of the tires.

Just thought I would add that in to further confuse some.

I'll bite - again (and hopefully I read it correct this time..)

The trailer has a center of gravity.. the distance to the rear wheels from the center of gravity hasn't changed any - I don't think the weight on the rear wheels changes any if it's supported by a tripod or the jacks..

Has nothing to do with the center of gravity. It has to do with the two points of support, the closer they get to each other the more weight will move from one to the other. That said that has nothing to do with getting the total weight by having the LG and axles on the same scale. Once you have the total weight then support the pin on the hitch, raise the LG, current weight now on the axles from total weight equals pin weight! Plain and simple.

And we finally have a winner:S

[rhagfo]

Vanished wrote:

bpounds wrote:

SoCalDesertRider wrote:
...
Now, on the pin weight versus jack weight issue, if you were weighing just the jacks by themselves, no trailer axles on the scale, that would not be the same as the pin, because there is about an 8 foot length difference between the pin to axle measurement, versus the jack to axle measurement. The jack weight will be significantly higher than the pin weight.

Interesting thing about this statement (which is true), is that it means the weight on your tires is less when your trailer is sitting at camp, than when you are hitched up. The weight forward of the landing gear is lifting weight off of the tires.

Just thought I would add that in to further confuse some.

I'll bite - again (and hopefully I read it correct this time..)

The trailer has a center of gravity.. the distance to the rear wheels from the center of gravity hasn't changed any - I don't think the weight on the rear wheels changes any if it's supported by a tripod or the jacks..

Has nothing to do with the center of gravity. It has to do with the two points of support, the closer they get to each other the more weight will move from one to the other. That said that has nothing to do with getting the total weight by having the LG and axles on the same scale. Once you have the total weight then support the pin on the hitch, raise the LG, current weight now on the axles from total weight equals pin weight! Plain and simple.

[Vanished]

rhagfo wrote:

Vanished wrote:
The trailer has a total weight.. In the OP's post example of 14,500..
He has two weights of 2900 (landing gear) and 11,600 (axles) as examples..
The question we don't know is where is the center of mass of the trailer.. The total weight of the trailer is the distance * force of the axles (11,600) plus the distance times force of the landing gear (2900).. the question we are trying to answer is what would the pin weight be? the answer is it also has to be 2900 lbs total - but it's further from the center of gravity - so if it's 16' from the center of mass compared to 12' for the jacks (both random examples) in theory it should be 2175 lbs if the examples above are used..

The OP's method can correctly calculate total shipping weight and axle weight but does indeed give an incorrect pin weight..

:S
You just don't get it do you?
Total trailer weight and on same scale pad gives total trailer weight. Lift landing gear and none of the TV is sitting on scale plate, now you get trailer axle weight, subtract that from total weigh is the True pin weight, plain and simple!

Read what I posted after I reread the OP.. I get it - Just misread it first..

[rhagfo]

Vanished wrote:
The trailer has a total weight.. In the OP's post example of 14,500..
He has two weights of 2900 (landing gear) and 11,600 (axles) as examples..
The question we don't know is where is the center of mass of the trailer.. The total weight of the trailer is the distance * force of the axles (11,600) plus the distance times force of the landing gear (2900).. the question we are trying to answer is what would the pin weight be? the answer is it also has to be 2900 lbs total - but it's further from the center of gravity - so if it's 16' from the center of mass compared to 12' for the jacks (both random examples) in theory it should be 2175 lbs if the examples above are used..

The OP's method can correctly calculate total shipping weight and axle weight but does indeed give an incorrect pin weight..

:S
You just don't get it do you?
Total trailer weight and on same scale pad gives total trailer weight. Lift landing gear and none of the TV is sitting on scale plate, now you get trailer axle weight, subtract that from total weight is the True pin weight, plain and simple!

[Vanished]

Sketch me up a free body diagram please - I'm a little slow today obviously.. 😉

[bpounds]

Vanished wrote:

I'll bite - again (and hopefully I read it correct this time..)

The trailer has a center of gravity.. the distance to the rear wheels from the center of gravity hasn't changed any - I don't think the weight on the rear wheels changes any if it's supported by a tripod or the jacks..

I'll give time to think about it. Or maybe someone else will chime in.

But here is a hint - this time there is a lever/fulcrum involved.