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Old 11-19-2005, 04:44 PM   #1
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Load distribution hitches-an analysis

For some years I have been thinking about some mathematical analysis to help understand how load distribution hitches operate. I’ve made my brain ache while considering dynamic behavior, including consideration of the different spring ratings of the various axles on the tow vehicle and trailer. After several fruitless attempts, I’ve decided to keep my analysis very simple, and only to examine the static situation in a non-rigorous manner. I’m going to make several incorrect assumptions, on the basis that they will only marginally affect the numerical results, while making a simple analysis possible, and some practical knowledge realizable. (For example, I assume the vehicles have no sprung or rubber suspension, so that the vehicles remain horizontal throughout) For those with little mathematical knowledge, rest assured that I will only be using the “balanced seesaw” method of calculating the forces in a static system. For example: Let us consider a small child of weight S, and a heavier child of weight B. They sit at opposite ends of a seesaw. For the seesaw to balance horizontally and remain level, the leverage (turning moment) exerted by each child must be equal. The turning moment of a force is equal to the magnitude of the force, multiplied by the distance of the weight from the turning point. To balance the seesaw, the heavier child, B, moves in towards the pivot, and at a distance L from the pivot, the seesaw balances. The turning moment of “child with weight B” about the pivot is thus B multiplied by the distance L, written as B*L. The other child, weight S, is sitting at a distance D from the pivot. This smaller child has a turning moment about the pivot of S multiplied by D, written as S*D. As the seesaw is balanced, the turning moments are equal and opposite in rotation, so we can write
B*L=S*D
If we divide both sides of this equation by B, using the symbol / to show division, we have
L=(S*D)/B
We can now answer the following type of question: How far (L) from the pivot of a seesaw does an 80 pound child (B) have to sit to balance a 50 pound child (S) who is 8 feet (D) from the pivot?. From the above equation, L= (50*8)/80 hence L=5 feet from the pivot.
We can now use this tool to consider a tow vehicle (TV) where:
W is the wheelbase of the tow vehicle (distance between front and rear axles)
H is the horizontal distance from the rear axle to the hitch ball
A travel trailer with tongue weight T is placed on the hitch. The back of the truck goes down, and the front goes up. The load distribution bars are left unconnected.
If we apply a downward corrective force, which we will call C, to the front axle of the tow vehicle, we can restore the front of the vehicle to its original position and loading. To calculate the value of C, we take turning moments about the rear axle of the tow truck:
C*W=T*H and hence
C= (T*H)/W , equation 1, giving the corrective force required to load down the front axle of the TV

The load distribution bars are now attached, and the chains adjusted until the front of the tow vehicle is lowered to its original height. That is, the bars are supplying the required downward force C at the front axle.
L is the length of the travel trailer from the hitch ball to the mid-point of its axle system. (In a triple axle, the middle axle, in a single axle, that axle, and in a double axle, the point half way between the axles.)
B is the length of the load distribution bars.
D is the total tension force in the load distribution chains.
Taking moments about the hitch ball for the trailer, the turning moment of the force D pulls down on the A frame, and this causes a compensating upward ground reaction A, at the travel trailer axles:
A*L=B*D and hence D= (A*L)/B (equation 2)
If, for the combined TV and trailer rig, we take moments about the rear axle of the TV:
C*W=A*(H+L) and hence A=(C*W)/(H+L) (equation 3), combined with equation 1, gives
A= (T*H)/(H+L) , equation 4, giving the load transferred to the trailer axles

The total weight of the rig remains unchanged, so the downward load increase at the front axle of the TV, C, and the downward loading of the trailer axles, A, must be balanced by a reduced loading, U, on the TV rear axle. Thus, U=A+C, and from equations 4 and 1, we have
U= ((T*H)/(H+L))+((T*H)/W) and hence
U= (T*H*(W+H+L))/(W*(H+L)) , equation 6, giving the reduced loading on the TV rear axle

From equations 2 and 4,
D=(C*W*L)/(B(H+L)) (equation 7)
From equations 7 and 1,
D= (((T*H)/W)*W*L)/(B(H+L)) and hence
D= (T*H*L)/(B(H+L)) ,equation 8, giving the total chain tension required

Let’s inject some real world figures into the equations. The approximate figures, guessing from memory, (forces in pounds, lengths in inches) for my trailer are:
T, tongue weight, = 800
H, rear overhang =60
L, trailer hitch to axle = 170
B, length of load bars = 33
W, tow vehicle wheelbase = 160
Inserting these values into the equations gives:
To restore the front level of the TV requires a total spring tension of 1075 pounds (537.5 in each chain). This will provide a download on the front axle of 300 pounds, a download on the trailer axles of 209 pounds, and a reduction in the TV rear axle loading of 509 pounds. The value of the extra load R on the rear axle, when the trailer with tongue weight T is hitched up without the chains, is obtained by taking moments about the front axle of the TV:
T*(W+H)=R*W and hence
R=(T*(W+H))/W , equation 9, giving the extra load on the rear axle before the chains are attached.
For my rig, R=1100 pounds
When the chains are attached, R is reduced by U, the uplift from the chain tension, so the final extra load E on the TV rear axle after the trailer is attached and the chains are tightened is given by:
E=R-U so from equations 9 and 6, after simplification,
E=(T*L)/(H+L) , equation 10, giving the final extra load on the TV rear axle, with chains tightened.
Inputting the values for my rig gives E=591 pounds

These equations help me to understand what happens when I connect my Excella to my Dodge truck. To summarize: When I lower the cup onto the ball, the 800 pound tongue weight loads the rear axle of the truck by 1100 pounds, and reduces the load on the front steering axle by 300 pounds. I then tighten the chains on the load distribution hitch, sufficient to reload the front axle to its original position and loading. The chains have to supply a tension of 1075 pounds to achieve this. The chain tension also reduces the rear axle load by 509 pounds from 1100 pounds to 591 pounds, and increases the travel trailer axle load by 209 pounds. I can obtain most of this information by going to a weigh station (which I have done), but the equations help the decision as to which strength of bars to purchase to fit to a given rig. Equation 8 gives the tension figure that can be given to the hitch manufacturer’s technical department. In words, the tension required is given by multiplying the tongue weight by the TV hitch rear overhang multiplied by the trailer length from ball to axle center, and dividing this figure by the product of the load bar length and the distance between the TV rear axle and the trailer axle center.
I chose just to restore the height of the TV front axle as my heavy duty truck is built to handle high loads on the rear axle. An uplift of the front axle is what I aim to avoid. If one wishes to add further loading to the front axle, a reasonable approximation would be to increase the chain tension in direct proportion to the desired increase. For example, if I wish to increase my front axle loading by 100 pounds from 300 to 400 pounds, an increase of 33%, I could increase the chain tension by 33% from 1075 pounds to 1433 pounds.
Interested readers could input into the equations the values for their rigs, or intended rigs. If there is sufficient interest, I could produce and e-mail a spreadsheet to do all the calculations, saving time otherwise spent with a calculator.
I would welcome comments on errors, omissions or improvements.
Nick.
.

Moderator Note: An Excel spreadsheet that performs these calculations has been posted at post 69 in this thread, http://www.airforums.com/forums/f464...tml#post475983
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Old 11-19-2005, 05:04 PM   #2
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ok

ummmm yeah, what he said.?
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Old 11-19-2005, 05:28 PM   #3
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It would seem to me that there must be an easier way for one to get to 500 posts
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Old 11-19-2005, 05:30 PM   #4
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Nick... this is an amazing post. You realize of course, that you lost me at "For some years"

Roger
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Old 11-19-2005, 05:32 PM   #5
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hmmmmmmmmmmmmmm.....maybe I should return to tent camping!!!!!
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Old 11-19-2005, 06:02 PM   #6
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I agree completely.

At the same time classical and central to modern mathematics, analysis involves studying continuous systems from dynamical systems to solutions of partial differential equations and spectra of operators.
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Old 11-19-2005, 06:14 PM   #7
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Come on people! This is simple linear equations - isn't it???
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Old 11-19-2005, 06:38 PM   #8
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B'Cup... are you familiar with MEGO syndrome? It hits me every time I see an "equals" sign...

Roger
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Old 11-19-2005, 06:46 PM   #9
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Nick,

Outstanding analysis. If someone would enter the spring rates (lbs/in deflection) for the various weight bars, we could combine this with Inland Andy's recommendation for 1-2 inch bend and tell exactly which bars to order.

I tried all you equations and they work very nicely.

Very good work!
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Old 11-19-2005, 06:54 PM   #10
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Um, er, I think this is why I chose a liberal arts major.
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Old 11-19-2005, 07:03 PM   #11
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Quote:
Originally Posted by 85MH325
B'Cup... are you familiar with MEGO syndrome? It hits me every time I see an "equals" sign...

Roger
Actually I do get into stuff like this. But what is even funnier is the wonderful responses to something like this.
What would happen if you combine this with the 4 point problem of the Hensley to come up with a complete analysis of the operation of that hitch...

Aw heck - I'll pass for now. I have a leak to fix...
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Old 11-19-2005, 07:36 PM   #12
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WHAT DID HE SAY? He is for it or against it. Boy am I lost. And I just got a new equalizer hitch and sway bar.
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Old 11-19-2005, 07:36 PM   #13
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U b a smart feller.....Makes PERFECT sense.........on a "static" model.


Now figure in "brake dive" and the loss of friction coefficient on a Reese Dual cam saddle as the pressure releases at the end of the bars. If you want to get fancy then figure in an evasive maneuver and the resulting body roll on the TV that will unload the bars at different rates.

That will once and for all put it to rest what bars with how much deflection is needed to make a DC work properly in an emergency situation and back up Inland Andy's observations about "over hitched".
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Old 11-19-2005, 08:11 PM   #14
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