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Old 09-16-2016, 05:09 PM   #1
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Weight distributing hitch question

I tow a 2016 Forest River TT with my 2010 Mercedes ML350 Turbodiesel. Trailer is exactly 7000 lbs, hitch weight 1070. (Yes, I am within the tow rating, and it tows very well). My question is really a math problem: When I tighten my equalizer hitch to distribute the weight back to the trailer and forward to the front wheels, isn't this INCREASING the torque load on the receiver hitch? It seems to me the receiver is acting as a fulcrum, and this acts in a twisting motion, basically trying to twist the receiver off its mount. (I have reinforced the receiver). It is probably very simple, but NOBODY seems able to answer the question, including the hitch person at the local Camping World and two local hitch specialists.
I get it that there is basically 1070 ft-lbs of torque (my ball is exactly one foot from the receiver). The question is, what happens to the torque exerted when I tighten the equalizer?
Any math experts out there?

Thanks.
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Old 09-16-2016, 05:37 PM   #2
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Well I'm no math expert but I don't think you are dealing with torque as torque deals with rotation. I think you are dealing with force. You have a downward force of 1070lbs onto your ball. This downward force is being distributed onto the rear of your TV and rear axle. Your rear axle is acting as the fulcrum and the distance from your ball to the axle is the is the lever raising your front wheels upward. This is the basic idea anyway. When you tighten your WD bars your are then exerting an upward force on your ball (by the spring force from the bars against your trailer frame) which removes some of the down force on the ball and transfers it to the front wheel... Anyway, that is how I understand it.

You should not be having to reinforce your receiver if it the right class for the job...
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Old 09-16-2016, 06:16 PM   #3
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Not sure how you can be within your rating when the max hitch weight is 600 lbs. for the ml350.
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Old 09-21-2016, 02:13 PM   #4
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You should not need to reinforce the receiver.


You are mixing apples and oranges a bit. Sounds like you understand the concept of torque being a load applied at a certain distance from a pivot point. But your 1070 TW has little to do with the torque exerted by the WDH system, which I believe would be much more than 1070ft-lbs.


When you put the trailer on the ball (no WDH engagement):
1070lbs is applied downward on the ball. You could say that this load is 1 foot from the receiver, but the WD head and receivers are plenty strong that they will not deform. So the load is transferred to the receiver.
Now the pivot in question is the attachments points of the receiver to the truck frame. The applied force is trying to pull the rearward bolts out of the frame, and push the forward bolts into the frame. But since they are sized properly, there is no relative movement, and the torque is transferred to the truck frame.
Now, the torque is pushing the rear axle downward, and lifting the front axle upward. Since the suspension is compliant, you see this movement. The springs react to the force, until equilibrium is achieved, and your rear end squats, and headlights point into oncoming traffic's eyes.


Solution? The WDH!!
You insert bars into the WD head, and you lift up on those bars. Forget the snap up chains or L-brackets that support the ends of the bars for now. Just imagine that you and a friend are holding the ends of the bars upward. Since this lifting is in the opposite direction of your 1070 TW, this lifting motion is directly counteracting the TW. You want to lift until the front axle is back where it was in the beginning. I don't know exactly how much load is required, because you are lifting both TW, truck weight, and trailer weight. Doesn't matter. But whatever the load, the torque is that load times the distance (I would say the distance to the front wheels as the pivot).


But you and your buddy can't run down the road to the campground holding those bars. Something else needs to support that load, and that's the trailer frame. When you place the bar-ends on the L-brackets, or the snap up chains for an old-style WDH, that load that you and buddy were holding is placed on the trailer frame. Some of that load is transferred back to the trailer wheels, but a good bit of it becomes additional TW. The math gets really tricky there, as it's a feedback loop. But rest assured, the entire rig will settle into equilibrium.


The result should be:
1. The front axle is back to it's original loading (stock ride height). If not, adjust the amount of WD.
2. The rear axle will have more load than original (some squat), but much less than without the WDH. Never allow the rear height to be greater than stock!
3. The trailer axles will have an insignificant amount of added load.
4. The trailer should be close to level. If not, adjust L-brackets and/or head height.
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Old 09-21-2016, 02:28 PM   #5
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Quote:
Originally Posted by thebrakeman View Post
You should not need to reinforce the receiver.


You are mixing apples and oranges a bit. Sounds like you understand the concept of torque being a load applied at a certain distance from a pivot point. But your 1070 TW has little to do with the torque exerted by the WDH system, which I believe would be much more than 1070ft-lbs.


When you put the trailer on the ball (no WDH engagement):
1070lbs is applied downward on the ball. You could say that this load is 1 foot from the receiver, but the WD head and receivers are plenty strong that they will not deform. So the load is transferred to the receiver.
Now the pivot in question is the attachments points of the receiver to the truck frame. The applied force is trying to pull the rearward bolts out of the frame, and push the forward bolts into the frame. But since they are sized properly, there is no relative movement, and the torque is transferred to the truck frame.
Now, the torque is pushing the rear axle downward, and lifting the front axle upward. Since the suspension is compliant, you see this movement. The springs react to the force, until equilibrium is achieved, and your rear end squats, and headlights point into oncoming traffic's eyes.


Solution? The WDH!!
You insert bars into the WD head, and you lift up on those bars. Forget the snap up chains or L-brackets that support the ends of the bars for now. Just imagine that you and a friend are holding the ends of the bars upward. Since this lifting is in the opposite direction of your 1070 TW, this lifting motion is directly counteracting the TW. You want to lift until the front axle is back where it was in the beginning. I don't know exactly how much load is required, because you are lifting both TW, truck weight, and trailer weight. Doesn't matter. But whatever the load, the torque is that load times the distance (I would say the distance to the front wheels as the pivot).


But you and your buddy can't run down the road to the campground holding those bars. Something else needs to support that load, and that's the trailer frame. When you place the bar-ends on the L-brackets, or the snap up chains for an old-style WDH, that load that you and buddy were holding is placed on the trailer frame. Some of that load is transferred back to the trailer wheels, but a good bit of it becomes additional TW. The math gets really tricky there, as it's a feedback loop. But rest assured, the entire rig will settle into equilibrium.


The result should be:
1. The front axle is back to it's original loading (stock ride height). If not, adjust the amount of WD.
2. The rear axle will have more load than original (some squat), but much less than without the WDH. Never allow the rear height to be greater than stock!
3. The trailer axles will have an insignificant amount of added load.
4. The trailer should be close to level. If not, adjust L-brackets and/or head height.
An eloquent explanation... thank you.
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