New Andersen WD hitch
A fellow camper bud showed me this. It's new and different. Anyone using one? Andersen WD hitches A U-tube video with the factory guy explaining it. You have to get past MR Truck doing his...
Forum Discussion
JBarca wrote:John, I don't know how Ed did it, but here is my approach to defining the relationship between tension and load transfer.
Hi Ed, Welcome to the discussion! When you came up with 1,000# per chain load, what are the assumptions surrounding this?
First we need to define some dimensions and variables. Let:
a = tow vehicle wheelbase
b = ball overhang (longitudinal distance from TV rear axle to ball)
c = distance from ball to mid-point between the TT's axles
d = perpendicular distance from Andersen chain to center of ball (reported by Andersen owner to be 6.5")
TW = tongue weight
LTT = load transferred to TT's axles
LF1 = load removed from TV's front axle due to TW without WD applied
LF2 = load transferred to TV's front axle when WD is activated
T = Andersen chain tension per chain
M = moment (torque) generated by Andersen chain tension (total for 2 chains)
then (assuming zero pitch-axis rotational friction between ball and coupler)
M = 2*d*T
LTT = M/c = 2*d*T/c
LF2 = LTT*(b+c)/a = 2*d*T*(b+c)/(a*c)
also
LF1 = TW*b/a
If we want to restore a load equal to some percentage (call it FALR) of that which was removed from the front axle, we have:
LF2 = LF1*FALR/100, or 2*d*T*(b+c)/(a*c) = FALR*TW*b/(a*100)
solving for chain tension (per chain) gives
T = FALR*TW*b*c/{2*d*(b+c)*100}
for example, if: b=60", c=200", d=6.5", TW=600#, and you want to restore 50% of the load removed (FALR=50)
T = 50*600*60*200/(2*6.5*260*100) = 1065# per chain
Ron
