Tim,
Thanks for your kind words and good questions. I will try to provide some good answers.
I guess I just don't see how the rear axle of the TV can serve as a "fulcrum" for ADDING weight to the front axle. Since the pivot point of these two rigid planes is the hitch ball, how can the fulcrum be anywhere other than at the hitch ball?
With a WD hitch, the hitch ball no longer is a pivot point because it is "bridged" by the WD bars and chains. The bars become a part of the structure and allow torques (moments) to be transmitted through the hitch assembly. The WD hitch acts as a rotational spring.
The way I see it, the weight added to the front axle should be derived by the moment equation, (300 lb)(200 in) = (____ lb)(195 in), or 308 pounds. Perhaps you could explain the error of my thinking here and explain how the rear axle of the TV is used as a fulcrum.
Static equilibrium requires that the sum of all external forces acting on a body is zero and that the sum of all external torques (moments) is zero. The summation of torques can be done using ANY point as a fulcrum or "pivot". We could use the ball coupler and the answers would be the same. The calculations would be more difficult.
We also could isolate the TV from the TT and calculate the TV axle loads. In this case the external forces are the two axle loads, the 2000# acting UP on the rear ends of the spring bars, and the 1700# acting DOWN on the ball. If we use either the front axle or the rear axle as the fulcrum, we can calculate the axle reactions directly. If we use the ball as the fulcrum, we would have to solve simultaneous equations.
Using the front axle as the fulcrum allows us to solve for the rear axle load (RAL) using the moment equation RAL x 130" = {2000# x (130"+65"+30" - (1700# x ( 130"+65")} giving RAL = (2000x225 -1700x195)/130 = 911.54#. This is the wheelbarrow analogy and the result is a "lifting" of the rear of the TV. IOW, a decrease in the rear axle load.
Using the rear axle as the fulcrum allows us to solve for the front axle load (FAL) using the moment equation FAL x 130" = {2000# x (65"+30") - {1700# x 65"}) giving FAL = (2000x95 - 1700x65)/130 = 611.54#. The 2000# with its longer lever arm overcomes the 1700# and tries to push the front of the TV down. IOW, the front axle load increases by 611.54#.
There's another thing that this thread has made me think about that I had never given consideration to before. That is the issue that arises when a dual axle trailer is not level, moving the virtual axle further forward or further back depending on whether the trailer is sitting front low or back low. I haven't done the math, but as contrary as it would first appear, I believe you could actually increase tongue weight by raising the tongue so the rear axle is supporting the bulk of the axle load. Of course, whether it can actually happen or not is dependent on how the axles are attached to the suspension and frame of the trailer.
Tandem axle trailers can have either "independent" suspension or "equalized" suspension. For either type, if you put the nose down, you move the TT's center of gravity forward slightly giving a slight increase in "tongue weight".
For the "independent" suspension, if you put the nose down, you increase the load on the front axle and decrease the load on the rear axle. This results in a decrease of "tongue weight". Depending on the spring constants, the decrease due to axle loads probably will be greater than the increase due to shifting the COG forward giving a net decrease in "tongue weight".
For the "equalized" suspension, putting the nose up or down does not cause any significant change in axle loads -- thats part of the purpose of the equalizer link. Therefore, there will be a slight increase in "tongue weight".
To quantify "slight", consider a TT which is 200" from ball to the axle fulcrum point. If the nose is dropped 2", the slope is 2/200 = 1%. If the TT's COG is 30" above the fulcrum point, the COG will shift forward about 30" x 1% = 0.3". If the TT has an equalized suspension, the change in spring angle might also shift the axles rearward by 0.1 to 0.2"
If the "tongue weight" percentage is 12% with the TT level, the COG initially will be 24" ahead of the fulcrum. With the nose down 2", the COG might be 24.5" ahead of the fulcrum. This changes the "tongue weight" percentage from 12% to 12.25%. It's hard to imagine how this small a change can affect "sway". If having the nose down does in fact improve TT stability, I think we should be looking for another explanation.
Thanks again for the good questions.
Ron
On edit: Tim, I forgot that I had discussed the "nose up/down" topic some time ago in
this post and the one following it.
