# Finding lbs per inch of a load

#### Shakermt

##### New member
So I had a guy show me a very cool method (which I forgot) to find the lbs/inch of a load. The reason is to find out how far to move the item on the trailer so it is balanced weight wise. I know the weight of the axle groups front and rear on the trailer and I know the distance from the kingpin to the center of the rear axle. I just can't for the life of me find the formula to find the lbs/in. I haul large machines and knowing the formula would save me massive amounts of time.

#### lev888

##### Senior Member
What is the exact calculation where the lb/inch quantity is used? A diagram and an explanation of terms like kingpin would help.

#### Shakermt

##### New member
Ok picture a semi trailer like the one below. The kingpin is where the trailer connects to the truck. When I scale the load out I will know the weights of each axle group. I also know the weight of the load +-500 lbs. I know the length of the machine as well.

#### Attachments

• 20210227_102618[1].jpg
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#### Cubist

##### Senior Member
I often like these real-world problems. But, that truck looks expensive therefore DISCLAIMER:- this post may be completely wrong therefore you must check that results are sensible and move the load slowly (especially the first time you try it!)

Code:
            load
|
|
V
K-------------------------------   trailer
^                      ^
|                      |
|                      |
Wf                     Wb

<---------------------->  ...distance D (between wheels)
<----->                   ...distance x (you don't have to measure this!)
K is the kingpin
Wf and Wb are the weights (forces) measured at the front/back wheels

Assume the trailer is light. Therefore Wf and Wb are zero when the trailer isn't loaded. If this isn't the case, then note down the un-loaded trailer's Wf and Wb and then...
Code:
  Unloaded Wf = 1200 lbs
Wf (due to the load) = 9200 - 1200
= 8000 lbs

Wb (due to the load) = 11000 - 1000
= 10000 lbs
Distance to move the load is

[MATH] D\times \left( \frac{1}{2} - \frac{W_b}{W_f + W_b} \right) [/MATH]
If this number is POSITIVE then move the load away from the kingpin.

-------------------

EXAMPLE. Trailer has D = 528 inches (distance between wheels). The wheel weights (due to the load) are measured at Wf= 8000 lbs and Wb = 10000 lbs

Distance to move the load =

[MATH] D\times \left( \frac{1}{2} - \frac{W_b}{W_f + W_b} \right) [/MATH]
[MATH] =528\times \left( \frac{1}{2} - \frac{10000 }{8000 + 10000} \right) [/MATH]
[MATH] =528\times \left( \frac{1}{2} - \frac{10000}{18000} \right) [/MATH]
[MATH]\approx -29 [/math] inches

This number is negative, therefore move the load toward the kingpin.

----------------------

THE MATH, if anyone is interested...
Code:
Take moments around the load

x*Wf = (D - x)*Wb

The above implies...

x = (D*Wb) / (Wf + Wb)

If weight is balanced, then x=x1 and Wf=Wb

x1 = (D*Wb) / (Wb + Wb)
x1 = D/2

When unbalanced, x=x2 and Wf ≠ Wb

x2 = (D*Wb) / (Wf + Wb)

= x1 - x2
= D/2 - (D*Wb)/(Wf + Wb)

Last edited:

#### Shakermt

##### New member
I often like these real-world problems. But, that truck looks expensive therefore DISCLAIMER:- this post may be completely wrong therefore you must check that results are sensible and move the load slowly (especially the first time you try it!)

Code:
            load
|
|
V
K-------------------------------   trailer
^                      ^
|                      |
|                      |
Wf                     Wb

<---------------------->  ...distance D (between wheels)
<----->                   ...distance x (you don't have to measure this!)
K is the kingpin
Wf and Wb are the weights (forces) measured at the front/back wheels

Assume the trailer is light. Therefore Wf and Wb are zero when the trailer isn't loaded. If this isn't the case, then note down the un-loaded trailer's Wf and Wb and then...
Code:
  Unloaded Wf = 1200 lbs
Wf (due to the load) = 9200 - 1200
= 8000 lbs

Wb (due to the load) = 11000 - 1000
= 10000 lbs
Distance to move the load is

[MATH] D\times \left( \frac{1}{2} - \frac{W_b}{W_f + W_b} \right) [/MATH]
If this number is POSITIVE then move the load away from the kingpin.

-------------------

EXAMPLE. Trailer has D = 528 inches (distance between wheels). The wheel weights (due to the load) are measured at Wf= 8000 lbs and Wb = 10000 lbs

Distance to move the load =

[MATH] D\times \left( \frac{1}{2} - \frac{W_b}{W_f + W_b} \right) [/MATH]
[MATH] =528\times \left( \frac{1}{2} - \frac{10000 }{8000 + 10000} \right) [/MATH]
[MATH] =528\times \left( \frac{1}{2} - \frac{10000}{18000} \right) [/MATH]
[MATH]\approx -29 [/math] inches

This number is negative, therefore move the load toward the kingpin.

----------------------

THE MATH, if anyone is interested...
Code:
Take moments around the load

x*Wf = (D - x)*Wb

The above implies...

x = (D*Wb) / (Wf + Wb)

If weight is balanced, then x=x1 and Wf=Wb

x1 = (D*Wb) / (Wb + Wb)
x1 = D/2

When unbalanced, x=x2 and Wf ≠ Wb

x2 = (D*Wb) / (Wf + Wb)

= x1 - x2
= D/2 - (D*Wb)/(Wf + Wb)

You sir rock. This will make life a lot easier when I am adjusting my load so I can balance out the weights.

#### Cubist

##### Senior Member
I've done some physical experiments to check the method above. As a result, there's an improved (more general) formula below. The same disclaimer applies to this!

It turns out that the trailer's unloaded weight plays a more important role than I had assumed in post#4. To cut a long story short, please ignore post#4 completely, the distance to move the load is actually...

[MATH] \frac{D(W_f - W_b)}{2(W_f + W_b - W_t)} [/MATH]
where $$W_t$$ is the trailer's unloaded weight (this amount can be obtained by adding together the front and rear axle's measured weight BEFORE the load is added).

EXAMPLE: Using the same data as above...
Loaded Wf = 9200 lbs, Wb = 11000 lbs
Unloaded front weight = 1200 lbs, unloaded rear = 1000 lbs, therefore trailer weight Wt = 1200+1000 = 2200 lbs
Trailer has length D = 528 inches (distance between wheels)
(NOTE: Don't do the post#4 step "adjust for trailer weight")

Distance to move the load =

[MATH] =\frac{D(W_f - W_b)}{2(W_f + W_b - W_t)} [/MATH]
[MATH] =\frac{528(9200 - 11000)}{2(9200 + 11000 - 2200)} [/MATH]
[MATH] =-26.4 [/MATH] inches. This number is negative, therefore move the load towards the kingpin. Note that the previous method is 2 and a half inches out (not a big deal, but there'd be a bigger difference for lighter loads or a heavier trailer).

Another advantage of this method is that it can be used to balance multiple loads. Just put all the loads on the trailer except the one that you plan to move around (for balancing). Then take a reading of the trailer's partially loaded weight (by measuring and adding the two axle weights). Call this quantity $$W_t$$. Now proceed as normal by adding the final load to the trailer, and using the above formula to determine how far to move the final load (leaving the others in place) to get balance. If you need to shuffle the initial load(s) to make space, then you don't need to re-weigh the partially loaded trailer (because the partially loaded weight, Wt, will remain the same). You'll just need to repeat the final, fully loaded, measurement which will tell you how far to move the final load (again) to compensate for the shuffling.

If you have a smartphone, you might be able to simplify the above procedure by creating a spreadsheet to perform the calculation (perhaps using Google Sheets).

Code:
           loadt  loadm
|      |
|      |
V      V
K-------------------------------   trailer
^                      ^
|                      |
|                      |
Wf                     Wb

<---------------------->  ...distance D (between wheels)
<------>                  ...distance xt (you don't have to measure this!)
<------------->           ...distance xm (you don't have to measure this!)

xm*Wf = (xm-xt)*Wt + (D-xm)*Wb

The above implies...

xm = (D*Wb - xt*Wt)/(Wb + Wf - Wt)

Find the unloaded trailer's xt and Wt

Wt = Wf + Wb
xt = (D*Wb) / (Wf + Wb)

If weight is balanced, then xm=xm1 and Wf1 = Wb1

xm1 = (D*Wb1 - xt*Wt)/(Wb1 + Wb1 - Wt)
xm1 = (D*Wb1 - xt*Wt)/(2*Wb1 - Wt)

When unbalanced, x=x2 and Wf2 ≠ Wb2

xm2 = (D*Wb2 - xt*Wt)/(Wb2 + Wf2 - Wt)

= xm1 - xm2
= (D*Wb1 - xt*Wt)/(2*Wb1 - Wt) - (D*Wb2 - xt*Wt)/(Wb2 + Wf2 - Wt)
Using Wb1 = (Wf2 + Wb2)/2
= (D*((Wf2 + Wb2)/2) - xt*Wt)/(2*((Wf2 + Wb2)/2) - Wt) - (D*Wb2 - xt*Wt)/(Wb2 + Wf2 - Wt)
= D*(Wf2 - Wb2)/(2*(Wb2 + Wf2 - Wt))

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#### Shakermt

##### New member
Wow, I have both the old and the new in a SS on my phone so I can use both.

#### Shakermt

##### New member
So real world test I used the new formula and I have included my 2 scale ticket screenshots. The first I was 3k heavy on my drives so I plugged the numbers into the spreadsheet and got 24 in. I moved it 22 (hard to see the measuring tape from 12' up lol ) and I was just 200 lbs off so thank you VERY much. You have made my life way easier.

#### Attachments

• Screenshot_20210719-191703_Weigh_My_Truck[1].jpg
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• Screenshot_20210720-060622_Weigh_My_Truck[1].jpg
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#### Cubist

##### Senior Member
So real world test I used the new formula and I have included my 2 scale ticket screenshots. The first I was 3k heavy on my drives so I plugged the numbers into the spreadsheet and got 24 in. I moved it 22 (hard to see the measuring tape from 12' up lol ) and I was just 200 lbs off so thank you VERY much. You have made my life way easier.

Thanks for posting these results! I'm very pleased that it worked. And wow, that's one HEAVY load We'll pop a freeMATHhelp.com bumper sticker into the post for your truck