WD Question For Ron

Wes Tausend wrote:
Roughly speaking, I once believed the axle(s) of a very short trailer, with a lengthy TV, can approach absorbing 2/3's tongue weight, and even a moderate TT + long WB truck, about 1/2 the tongue weight to the TT axle, if both truck axle springs are compressed equally. But this doesn't seem to quite compute with Ron's formula above, if I got it right. It seems the distance from ball to front axle must come into play. I think Ron's formula is correct but may not be applicable to the this question.

Wes, if you are referring to the formula in this post, you need to be aware that formula is for the special case of returning the front axle to its unhitched load.

Your particular scenario, if I interpret correctly, pertains to how much of the "tongue weight" ends up on the front axle, rear axle, and TT axles after application of WD.
For this case, you first must calculate how much load is removed from or added to the various axles due to application of tongue weight.
Then you must calculate how much load is transferred to/from the axles due to application of WD.

Let:
Lf = Net load added to TV's front axle
Lr = Net load added to TV's rear axle
Ltt = Load transferred to TT's axles
D1 = TV's wheelbase
D2 = Distance from TV's rear axle to ball
D3 = Distance from ball to midpoint between TT's axles

For your condition of Ltt = (2/3)*TW,
Lf/TW = (2*D3-D2)/(3*D1)
Lr/TW = (D1+D2-2*D3)/(3*D1)

Picking some convenient example dimensions: D1=130", D2=65", and D3=195" gives --
Lf/TW = +83%
Lr/TW = -50%

If TW=900#, the net axle load changes are: 750# is added to the front axle,
450# is removed from the rear axle, and 600# is added to the TT axles.
And, you would need WD bars rated for about 2000# each.

If you changed D3 to 97.5" (a very short trailer), you could have 1/3 of the TW added to the front axle, zero added to the rear axle, and 2/3 of TW added to the TT axles.

Ron