New Andersen WD hitch

Ron Gratz wrote:

JBarca wrote:
Hi Ed, Welcome to the discussion! When you came up with 1,000# per chain load, what are the assumptions surrounding this?

John, I don't know how Ed did it, but here is my approach to defining the relationship between tension and load transfer.

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

Hi Ron,

Catching back up on this.

I used your formula,

Ron G wrote:
solving for chain tension (per chain) gives

T = FALR*TW*b*c/{2*d*(b+c)*100}

And check it against my assumptions with my TT & TT.

I my case I used a 1,400# TW, 90% FALR and at the time I estimated "d" to be 8" where you used 6.5" So I tweaked the formula to use 8 to see if it came close to how I backed into it.

Both methods came in close to similar areas.

Using your formula gave 4,050# per chain.

Using my method 3,919# per chain or 131# less than you did.

From here:

JBarca wrote:
In my case I have an actual 1,400# TW and this hitch is rated that high. That is approx. 1,100# at each snap up chain using 28.5” long WD bar or 5,225 ft. lb of torque into the receiver. The Anderson using it’s 8” tow ball would need 7,838# total chain force to create this.

There is some error in this as my Reese Tow beast shank is longer then the Anderson shank but it get’s me in the league of feeling what chain forces are going on with the Anderson.

The 28.5” WD bar compared to the 8” long tow ball is the difference in the mechanical advantage of the 2 hitches for WD. The Anderson will always have to use higher chain force to get the same WD.

I used the spread sheet you sent me many years ago on backing into the chain force of a traditional WD hitch for a given TV and TT. Once you have the chain force you can create the torque in the receiver to the amount of weight returned to the front of the TV. Since I had scale weights and know how my hitch was adjusted, I figured out the torque in the receiver with the Reese then put the Andersen in the same setup to reproduce the same torque, thus creating a chain tension in the Andersen.

Point: 2 approaches that come very close to the same answer.

Also , thanks for the link to the AS site. I read that a while ago but did not follow the latest info.