All WD systems need a minimum amount of tongue weight in order to work properly. Call a few companies like Lindon Hitch and they will tell you that. That is because you need that minimum weight at the ball in order for the bars to distribute it. ---
And what would the magnitude of that "minimum weight" be? 500#? 50#? 5#? How is it determined? Can you provide the name of someone who told you this? I certainly would like to communicate with that person. Perhaps you are thinking of the Dual Cam's requirement for a certain amount of WD bar load in order to make the SWAY CONTROL work properly.
--- If you have 0 tongue weight, what weight can you expect to distribute?
I can expect to distribute some of the load which is on the TV's rear axle. In this case, it would be load resulting solely from the weight of the TV. This would be similar to the way a liftable tag axle on a concrete hauler redistibutes the axle loads when it is lowered.
Furthermore, your original drawing shows: TV receiver load WD hitch REMOVES 300#'s.
1. Where is this load removed from ? ---
The WD hitch produces an UP force of 2000# on the hitch head and a DOWN force of 1700# on the ball coupler. The net result is an UP force (removal) of 300# on the receiver. If the tongue weight were zero to begin with, the hitch now would be pushing UP with a force of 300# on the receiver. Whether the receiver load goes from 300# DOWN to zero or goes from zero to 300# UP makes no difference to the TV.
2. You are saying in your drawing: There is 300#s removed from the TV's receiver, but then you claim that the load on the TV's ball which is directly connected to the reciever and is directly under and connected to the tongue remains the same. How is this possible? It seems you are disagreeing with yourself.
I have never said that the "load on the TV's ball --- remains the same". I have said more than once that the load on the ball increases by 1700# using my example WD hitch application.
Let me try once more:
Begin with X load on the ball which produces X load on the receiver.
Use the WD bars to pull down on the A-frame with 2000# which causes a DOWN force of X+1700# on the ball coupler.
You now have X+1700# acting down on the ball and hitch head.
Simultaneously, the front end of the WD bars push UP with a force of 2000# on the hitch head.
The net result (receiver load after WD minus receiver load before WD) is (X+1700-2000)-X = -300#. The WD has removed 300# from the receiver and it does not make any difference what the value of X is.
Thomas, since I invited questions and comments, I feel obligated to continue this as long as you want to. I just ask you again to please compare your definition of "tongue weight" with my definition of "tongue weight". We seem to be trying to compare apples and oranges.
Ron
