The Andersen WD Hitch User Thread

[mstephens]

Improved Calculation

I have attached the improved calculation of chain strain thanks to Ron Gratz, who was kind enough to point out my original error, AND supply the correction. I therefore dubbed this drawing the "Ron Gratz Method."

I think it is fascinating to see exactly how each variable affects the final result. It is easy to see how TV wheelbase and overhang affect the result, and the TT ball to axle dimension. The math only requires some basic arithmetic, so anyone can do it.

No, this is NOT instructions on "how to setup your hitch." You get those instructions from the hitch manufacturer. This is merely an explanation of forces to provide an understanding of how they will change depending on your TV/TT dimensions and how much weight you expect to transfer.

The calculations here take the ball center to chain plate dimension as "6.5" which was reported by Bruce. This is a crucial dimension because of it's effect on leverage. Since I didn't have Bruce's TT wheelbase, I just plugged in a sample of 150".

EXPLANATION
Any downward force on the TV front axle must be countered by an upward force on the TT axle in order to have equilibrium. Using the TV rear axle as a fulcrum (noted in drawing), the ratio of the two distances determines how much upward force is needed at the TT axle to counter the downward force at the TV front axle. In the EXAMPLE a 200# weight shift to the front of the TV requires a 121# upward force at the TT rear axle.

Once the upward force is known - 121# in this case - the remaining calculation computes the amount of torque must be applied to the ball/coupler to equal the 121# upward force. What is crucial here is the ratio of the long lever to the short one. The long one is the ball to TT axle, and the short one is the ball to chain plate. As the ratio increase, the force required to torque the ball gets higher.

By looking at the formulas, you can see that strain on the chains will increase when: The weight to be shifted increases, or the TV wheelbase increases, or when the overhang decreases, or when the TT ball to axle decreases. These are all linear equations, so the effects are proportional on each variable. A 10% increase in TV wheelbase creates a 10% increase in strain. Likewise then, a 20% increase in weight shift will create a 20% increase in strain, and so on.

This is just for information. Some people, like me, enjoy knowing "how things work." This is not an instruction on how to setup your hitch, which should be done by using the manufacturer's instructions.

[mstephens]

Bruce--
Ok, 164". I'll see if I can correlate to your weight and scale measurements.

[crisen]

I sent this to Bruce H yesterday and have revised to it based on his number of 164" on the length of the trailer ball to axle.

This is my take on what is happening. I am an engineer but haven't spent much time on the problem so don't hold it against me. This explaination takes a little time so bear with me.

I think you have to calculate how much moment is applied to the hitch reciever to remove the required weight of on the rear axle then this weight is distributed to the front axle and trailer axle, inverserly proportional to their distance from the rear axle of the tow vehicle (by that I mean the closer to the rear axle the axle under consideration is the more weight it gets in the weight transfer) . Or put another way the WD hitch removes weight from the rear axle based on the moment generated and the weight removed is transferred to the trailer axle(s) and front TV axle due to the effects of weight transfer from raising the back of vehicle. This latter weight transfer is similiar to weight jacking on a race car.

I found this first formula in a book to calculate the weight added to the rear axle of a tow vehicle by a trailer. I went thru it with your numbers to check it.

Rear axle weight added = (dist ball to rear axle/wheel base x TW) + TW

I believe you said the Honda was 155" ball to front axle and Honda says wheel base is 106.3", also a post said TW was 400 lb.

R.A. W. added = (48.7/106.3 x 400) + 400 = 583lb
Your chart says 600 lb so this checks.

Now to calculate the chain forces and axle loads.

Now your chart says you removed 260 lb from rear axle with 3/16 compression of the bushing. The moment at the ball to do this is 260 lb x 48.7 inches = 12,662 in-lb.

This moment is generated by the pull on the chains x the ball to pivot distance which you stated is 6.5 inches.

Force on 2 chains = 12,662 in-lb/6.5 inches = 1948 pounds.

1948 total/2 chains = 974 lb pull per chain

You measured 970 lb on the Sherline if I remember correctly, so that checks.

Now the front axle and trailer axle have to get the 260 lb removed from the rear axle.

I don't have the dimensions of the trailer but your data says there were 180 added to the Honda front and a post said there were 76 added to the trailer. 180 + 76 = 256 or pretty darned close to the 260 removed from the rear.
(Todays addition)
Recently you posted ball to trailer axle distance at 164" so calculate the weight transfer as follows. Remember the transfer is inverse to the distance to the axle when reading these numbers.

Weight added to trailer axle = (W.B of TV/ total distance TV front to trailer axle) x weight tranfered from rear axle = (106.3/106.3+48.7+164) x 260 lbs = 86.7 lb.

I think your post said trailer axle went from 3044 to 3120 = 76 lb

By the same token the weight added to the front axle is (TV Rear axle to trailer axle/total distance front axle to trailer axle) x weight remove from rear axle = ((164+48.7)/106.3+48.7+164) x 260 = 173 lb

Your table shows 180 lbs.

Anyway that is my take.

[Bruce H.]

Crisen,

WOW! Like I said in my private response to you last night, I am amazed at the congruence between your predicted weight distribution and the actual weight measurements.

Thank you so much for your research.

Bruce

[MarkR]

Originally Posted by mstephens

Improved Calculation
EXPLANATION
Any downward force on the TV front axle must be countered by an upward force on the TT axle in order to have equilibrium. Using the TV rear axle as a fulcrum (noted in drawing), the ratio of the two distances determines how much upward force is needed at the TT axle to counter the downward force at the TV front axle. In the EXAMPLE a 200# weight shift to the front of the TV requires a 121# upward force at the TT rear axle.

I hope to need a WD Hitch in about 8 months (my new deadline for finishing my trailer). I have appreciated everyone's posts, AND the original "poster", AND want to officially thank everyone for doing so (AND the Mods for keeping it open). Short of some unforeseen event, I will likely purchase this Hitch, seems like it will be a good match for my needs.

Disclaimer: I'm not an engineer, but sometimes pretend to be. Is there a way to manipulate the formula so I can use my "likely" Tongue Weight to find out how much strain will be in the Chain in order to equalize the diagram - seems like it would be possible since I know all the other variables/dimensions/weights . . . just don't know the TV axle weights. And I would be wanting to do this for kicks - to play w/the free body diagram - and nothing else.

Thanks again,
MarkR

[mstephens]

Originally Posted by MarkR

Disclaimer: I'm not an engineer, but sometimes pretend to be. Is there a way to manipulate the formula so I can use my "likely" Tongue Weight to find out how much strain will be in the Chain in order to equalize the diagram - seems like it would be possible since I know all the other variables/dimensions/weights . . . just don't know the TV axle weights. And I would be wanting to do this for kicks - to play w/the free body diagram - and nothing else.

Thanks again,
MarkR

Mark,
You will need to know how much weight you want to transfer to the front wheels, before you can calculate the chain strain. It is the transferring which requires force through the chain and hitch. Just knowing the TW isn't enough.

Looking at Bruce's Figures
I came up with a slightly higher chain strain.

W = 180#
Wheel Base = 106"
Overhang = 49"
Hitch to TT axle = 164"
Ball to chain plate = 6.5"

Then...

F = 180 x 106/213=89
S= 89 x (164/6.5)/2= 1,122 lbs. Clearly higher than the gauge measure.

[Wazbro]

Originally Posted by mstephens

Improved Calculation

EXPLANATION
Any downward force on the TV front axle must be countered by an upward force on the TT axle in order to have equilibrium. Using the TV rear axle as a fulcrum (noted in drawing), the ratio of the two distances determines how much upward force is needed at the TT axle to counter the downward force at the TV front axle. In the EXAMPLE a 200# weight shift to the front of the TV requires a 121# upward force at the TT rear axle.

A WD hitch adds weight to the TV front axle and the TT axles so wouldn't it be a downward force on both?

[idroba]

Gosh, a bumblebee can fly after all. Big grin.

I believe one other factor may come into the calculations. If you have a drop shank setup, I think the effective distance (I believe the calculations used 6.5 inches) is different depending on the bottom plate to hitch receiver distance, not the ball center. Thus, if you were pulling (twisting) from a longer drop you effectively have more lifting on the TV frame.

But, sitting here on the top of Marias pass, highway 2, just below Glacier National Park, towing with my Andersen, it all seems kinda academic. The hitch does work well. BTW, some white lube or grease on the frame tube ends does seem to help with the chain wear, but time will tell with that. You guys figure out how to solve that issue next please.

I will enjoy towing and camping while it is all worked out.

[purman]

Originally Posted by idroba

Gosh, a bumblebee can fly after all. Big grin.

I believe one other factor may come into the calculations. If you have a drop shank setup, I think the effective distance (I believe the calculations used 6.5 inches) is different depending on the bottom plate to hitch receiver distance, not the ball center. Thus, if you were pulling (twisting) from a longer drop you effectively have more lifting on the TV frame.

But, sitting here on the top of Marias pass, highway 2, just below Glacier National Park, towing with my Andersen, it all seems kinda academic. The hitch does work well. BTW, some white lube or grease on the frame tube ends does seem to help with the chain wear, but time will tell with that. You guys figure out how to solve that issue next please.

I will enjoy towing and camping while it is all worked out.

Well said , a 100 or so post of numbers that were wrong and a few more that are right, really seem enough for a supposedly users thread IMO. I'm glad everyone got their number fix now.

Back to the issue that needs to have some posts about it.. CHAIN WEAR.

It's been raining here for the last 2 days so I haven't had a chance to figure out my chain wear yet. hopefully tomorrow I will get a chance to look at it.

My plan is to move the brackets so the chain is in a direct line with the plate. We shall see how it looks. Then I have to decide what method I want to use to keep the bracket from moving. decisions, decisions...

[mstephens]

Originally Posted by Wazbro

A WD hitch adds weight to the TV front axle and the TT axles so wouldn't it be a downward force on both?

To SHIFT weight, a downward force has to be countered by an upward force in order to keep the combined weight in equilibrium.

e.g. If your total rig weighs 4,000#, and you shift 300# to the front axle, the total rig will still weigh 4,000#. What is ADDED to one axle, must be taken off other axles.

[purman]

There are Threads on Weight distribution. PLEASE lets try and stay on topic. "The Andersen WD Hitch User Thread, I believe is the thread title.

[mstephens]

Here's how I used the calculations to see if this ANDERSON hitch would work for me.
My rig has an 800# tongue, and I need to shift about 300# to the front axle.
My measurements TV/TT measurements are:

W to shift = 300#
TV wheelbase = 132"
TV overhang = 60"
Ball to TV axle center (2 axles) = 192"
Ball center to chain plate = 6.5"

F = 300 x 132/252 = 157 lbs
S= 157 x 192/6.5/2 = 2,318 lbs on each chain. That's over the WLL of the standard 2000# chain on the ANDERSON HITCH. So, I could not safely get a 300# weight transfer using the standard chain.

If I cut back to a 250# transfer...
F= 250 x 132/252 = 130 lbs
S= 130 x 192/6.5/2 = 1,920 lbs on each chain. Under the 2,000# limit, but barely. Maybe I would want to limit the pull to say, 1,700# per chain.
That would give me an allowable weight shift of ~215 lbs onto the front axle.

For me, that simple pencil and paper exercise is a little easier as a rough evaluation than buying the ANDERSON HITCH, taking off my old unit, mounting the new ANDERSON HITCH, doing a lot of weigh-ins, and then possibly not knowing I was at the WLL of the chains. Knowing what I know from the pencil and paper, I would now inquire about heavier chains for the ANDERSON HITCH. Which, as a potential user of the ANDERSON HITCH, I will do.

[purman]

Originally Posted by mstephens

Here's how I used the calculations to see if this ANDERSON hitch would work for me.
My rig has an 800# tongue, and I need to shift about 300# to the front axle.
My measurements TV/TT measurements are:

W to shift = 300#
TV wheelbase = 132"
TV overhang = 60"
Ball to TV axle center (2 axles) = 192"
Ball center to chain plate = 6.5"

F = 300 x 132/252 = 157 lbs
S= 157 x 192/6.5/2 = 2,318 lbs on each chain. That's over the WLL of the standard 2000# chain on the ANDERSON HITCH. So, I could not safely get a 300# weight transfer using the standard chain.

If I cut back to a 250# transfer...
F= 250 x 132/252 = 130 lbs
S= 130 x 192/6.5/2 = 1,920 lbs on each chain. Under the 2,000# limit, but barely. Maybe I would want to limit the pull to say, 1,700# per chain.
That would give me an allowable weight shift of ~215 lbs onto the front axle.

For me, that simple pencil and paper exercise is a little easier as a rough evaluation than buying the ANDERSON HITCH, taking off my old unit, mounting the new ANDERSON HITCH, doing a lot of weigh-ins, and then possibly not knowing I was at the WLL of the chains. Knowing what I know from the pencil and paper, I would now inquire about heavier chains for the ANDERSON HITCH. Which, as a potential user of the ANDERSON HITCH, I will do.

Just a note:
They are rated to 2000# but chains are rated at 1/3 of the braking strength. Which means they won't brake till they hit around 6000#

Reason: most people who use chains don't worry about the load they put on them. ie. Logging, trucking, ships, etc. But it would be interesting to see if Andersen will make them with heavier chains.

Another consideration is where the chain is welded to the bolt. I would think this weld would fail before the chain would? Or the clip and pin that hold the chain to the plate. IMO

[xrvr]

on the defensive now, please give us more theory. Seat of the pants experience trumps theory always no matter what the experts say. Jim

[purman]

Ok maybe not the weld that thing is beefy.

[purman]

Originally Posted by zigzagguzzi

on the defensive now, please give us more theory. Seat of the pants experience trumps theory always no matter what the experts say. Jim

Not sure what you are trying to say. But yes real world experience usually does trump theory from behind a desk.

[Wazbro]

Originally Posted by mstephens

To SHIFT weight, a downward force has to be countered by an upward force in order to keep the combined weight in equilibrium.

e.g. If your total rig weighs 4,000#, and you shift 300# to the front axle, the total rig will still weigh 4,000#. What is ADDED to one axle, must be taken off other axles.

Sorry fact is weight comes off rear axle to go to front axle and trailer axle check weight scale measurement threads.

[SteveH]

I believe the weak point will not be the chains or the welds, but the shackles that connect the chain to the plate.

[purman]

Originally Posted by SteveH

I believe the weak point will not be the chains or the welds, but the shackles that connect the chain to the plate.

This is my thought. It doesn't have a weight listed on it. ? When I go to the hardware store I will see what similar size ones are rated too. The Shackle I put in to extend my chain is rated to 2800#

[Ron Gratz]

Originally Posted by mstephens

EXPLANATION
Any downward force on the TV front axle must be countered by an upward force on the TT axle in order to have equilibrium. Using the TV rear axle as a fulcrum (noted in drawing), the ratio of the two distances determines how much upward force is needed at the TT axle to counter the downward force at the TV front axle. In the EXAMPLE a 200# weight shift to the front of the TV requires a 121# upward force at the TT rear axle.

Mark, the WDH causes load to be added to the TV's front axle and to the TT's axle pair. A load equal to the sum of these is removed from the TV's rear axle. This could be shown by downward forces acting on the front and TT axles and an upward force acting on the rear axle. Showing the front axle force and TT axle force acting in opposite directions is not correct.

Also, in the equations on your sketch, the product of a load or weight multiplied by a lever arm would be a torque rather than a force.

Ron