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 > Adjusting Equalizer hitch for new TV

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padave

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Posted: 07/01/12 05:37am Link  |  Quote  |  Print  |  Notify Moderator

Got a new TV....broke down and bought an Ecoboost F150. Question, when adjusting, how do you know how many spacer washers you need? Going from a Titan SWB of 140" to F150 with 157" wheelbase. Do you think that I will need to add a washer or two?


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pulsar

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Posted: 07/01/12 06:23am Link  |  Quote  |  Print  |  Notify Moderator

Moved from Forum Technical Support.

BarneyS

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Posted: 07/01/12 07:57am Link  |  Quote  |  Print  |  Notify Moderator

The only way to know is to hook up, measure or weigh, and see. I suspect that it will remain just about the same.
Barney


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jmtandem

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Posted: 07/01/12 08:19am Link  |  Quote  |  Print  |  Notify Moderator

Quote:

Going from a Titan SWB of 140" to F150 with 157" wheelbase. Do you think that I will need to add a washer or two?


The wheelbase of the truck has nothing to do with adjusting the hitch. The height of the receiver between the two trucks would be the difference in hitch head height on the shank that might need adjustment. Be certain the trailer is sitting level fully loaded for camping and the spring bars are nearly parallel to the trailers A frame correctly tensioned. Approximately one washer in the head equals one hole in the L brackets for adjustment.

* This post was edited 07/01/12 08:29am by jmtandem *


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Posted: 07/01/12 09:02am Link  |  Quote  |  Print  |  Notify Moderator

Starts on page 17... Equal-i-zer Owner's Manual


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mkirsch

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Posted: 07/01/12 09:02am Link  |  Quote  |  Print  |  Notify Moderator

There's no way to know what you need until you try it. The whole process is a bit of trial and error.


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Posted: 07/01/12 10:39am Link  |  Quote  |  Print  |  Notify Moderator

padave wrote:

Got a new TV....broke down and bought an Ecoboost F150. Question, when adjusting, how do you know how many spacer washers you need? Going from a Titan SWB of 140" to F150 with 157" wheelbase. Do you think that I will need to add a washer or two?
I agree with jmtandem that a change in TV wheelbase should not require a change in WDH ball mount tilt or L-bracket height adjustment.

The required load transfer is slightly dependent on distance from TV's rear axle to tow ball.
But, there's probably not enough difference in that parameter, for the two TVs, to make any difference in the WDH adjustment.

If the OP wants to follow Ford's new specifications for WDH adjustment, he might need to remove a washer or two.

Ron

Searching_Ut

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Posted: 07/01/12 11:52am Link  |  Quote  |  Print  |  Notify Moderator

Wheelbase most definately come in to play. We're talking leverage, and distance on both sides of the pivot point are part of the equation. Just follow the directions to which a link has already been provided and I think you'll find it's fairly easy to set up, although you may have to try several combinations to get it just right.

Chuck&Gail

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Posted: 07/01/12 02:02pm Link  |  Quote  |  Print  |  Notify Moderator

The only RIGHT way is to follow the directions. Every vehicle is different.


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Posted: 07/01/12 02:05pm Link  |  Quote  |  Print  |  Notify Moderator

Searching_Ut wrote:

Wheelbase most definately come in to play. We're talking leverage, and distance on both sides of the pivot point are part of the equation.---
Since you mentioned equations --

When the coupler is placed on the ball, the load removed from the TV's front axle is approximately: L = TW*OH/WB
where TW = tongue weight, OH is ball overhang (distance from TV's rear axle to ball), and WB = TV's wheelbase.

For the Titan: L1 = TW*OH1/140
For the F-150: L2 = TW*OH2/157

Then we have to select a criterion for adjusting the WDH. Let's assume we want to adjust the WDH so it restores 100% of the load which was removed from the front axle.
The load restoration is approximately: R = WDF*BL/TTD*(OH+TTD)/WB
where WDF is the combined downward force exerted by the WD bars against the A-frame, BL is the effective length of the WD bars, and TTD is the distance from ball to midpoint between TT's axles.

For the Titan: R1 = WDF1*BL/TTD*(OH1+TTD)/140
For the F-150: R2 = WDF2*BL/TTD*(OH2+TTD)/157

Since we assume the restored load will be equal to the removed load, we have R1 = L1 and R2 = L2. That gives
WDF1*BL/TTD*(OH1+TTD)/140 = TW*OH1/140, and
WDF2*BL/TTD*(OH2+TTD)/157 = TW*OH2/157

A little bit of algebra simplifies the above to:
WDF2/WDF1 = {OH2/OH1}*{(OH1+TTD)/(OH2+TTD)}

If you replace OH1, OH2, and TTD with any realistic values for the OP's TVs and TT, you should see that there is relatively little difference in the theoretical values of WD bar force required for the two TV's. And, what difference there is, does not depend on TV wheelbase.

Ron

* This post was edited 07/01/12 02:50pm by Ron Gratz *

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