Weight distributing hitch math?

That's the very "Toolbox" I'm trying to learn to use, Westend!

Thanks to all for the tips/information. I'm getting a lot clearer picture all the time.

I think I see what you're after, Francesca, the amount of force at the intersection of the tongue and the frame, no?
To find the amount of force that should be OK with the materials of the tongue, frame, and attachment will be an engineering exercise. Engineering Toolbox Also, the forces will be dynamic when tug and trailer are in motion. You should be able to get to a good ballpark maximum of those, though.

Francesca Knowles wrote:
I understood that tongue weight remains the same- was just asking if that (original) number is the one you refer to in your statement re. pounds of force (below).

Ron Gratz wrote:
The second bar also will exert the same force on its A-frame. Total downward force on the "tongue" will be approximately twice the tongue weight.

"Tongue weight" is the tongue weight of the loaded trailer.

There are two WD bars. Each bar exerts a force approximately equal to the tongue weight.

Assuming the "tongue" is comprised of two beams (forming an A-frame) ,
one WD bar exerts a downward force on the left side of the A-frame,
and the other bar exerts a downward force on the right side of the A-frame.

Ron

Ron Gratz wrote:

Francesca Knowles wrote:
As used above, does "tongue weight" refer to its original weight before W/D is added?

If it's correct to assume that force to be concentrated at the attachment point on the frame, can/would it be expressed as so-many-pounds-per...something?

Tongue weight remains the same -- it is not changed by the application of WD.

The force exerted on the A-frame is expressed in pounds -- just as tongue weight is expressed in pounds.

Ron

I understood that tongue weight remains the same- was just asking if that (original) number is the one you refer to in your statement re. pounds of force (below).

Ron Gratz wrote:
The second bar also will exert the same force on its A-frame. Total downward force on the "tongue" will be approximately twice the tongue weight.

Francesca Knowles wrote:
As used above, does "tongue weight" refer to its original weight before W/D is added?

If it's correct to assume that force to be concentrated at the attachment point on the frame, can/would it be expressed as so-many-pounds-per...something?

Tongue weight remains the same -- it is not changed by the application of WD.

The force exerted on the A-frame is expressed in pounds -- just as tongue weight is expressed in pounds.

Ron

Ron Gratz wrote:
The second bar also will exert the same force on its A-frame. Total downward force on the "tongue" will be approximately twice the tongue weight.

Dawn may actually be breaking in my brain...

As used above, does "tongue weight" refer to its original weight before W/D is added?

Also:

If it's correct to assume that force to be concentrated at the attachment point on the frame, can/would it be expressed as so-many-pounds-per...something?

Francesca Knowles wrote:

Ron Gratz wrote:

Francesca Knowles wrote:
Can the stress on the trailer's tongue at the "red arrow" point be expressed/predicted in terms of pounds?

Yes, depending on TV/TT dimensions --
if you want to restore all of the load which was removed from the front axle,
the force applied to the A-frame by a single WD bar will be approximately equal to the tongue weight.

Then, if you know how to do beam bending and shear stress calculations, you can calculate how much stress will be induced in the A-frame.

Ron

In the above example, does the second bar also exert force equal to tongue weight, for a total of twice the tongue weight at the a-frame?

Also, per calculations: is the W/D a first, second, or third class lever? The "cheater page" I found thinks I already know that so I'm not gettin' anywhere until I find out...

The second bar also will exert the same force on its A-frame. Total downward force on the "tongue" will be approximately twice the tongue weight.

The TT and tongue act as a Class 2 lever. Either the TT's axles or the ball coupler can act as the fulcrum, and the load (downward force from the chains) is applied in between.

Ron Gratz wrote:

Francesca Knowles wrote:
Can the stress on the trailer's tongue at the "red arrow" point be expressed/predicted in terms of pounds?

Yes, depending on TV/TT dimensions --
if you want to restore all of the load which was removed from the front axle,
the force applied to the A-frame by a single WD bar will be approximately equal to the tongue weight.

Then, if you know how to do beam bending and shear stress calculations, you can calculate how much stress will be induced in the A-frame.

Ron

In the above example, does the second bar also exert force equal to tongue weight, for a total of twice the tongue weight at the a-frame?

Also, per calculations: is the W/D a first, second, or third class lever? The "cheater page" I found thinks I already know that so I'm not gettin' anywhere until I find out...

Thanks for bearing with me.

Francesca Knowles wrote:
Can the stress on the trailer's tongue at the "red arrow" point be expressed/predicted in terms of pounds?

Yes, depending on TV/TT dimensions --
if you want to restore all of the load which was removed from the front axle,
the force applied to the A-frame by a single WD bar will be approximately equal to the tongue weight.

Then, if you know how to do beam bending and shear stress calculations, you can calculate how much stress will be induced in the A-frame.

Ron

Francesca Knowles wrote:
Thing is, I keep looking at pictures, and it appears to me that the bars/chains must be pulling on the trailer tongue at the connection point, pushing it down against the fulcrum provided by the ball. I s'pose that's wrong, but it's the only logic my simple mind can come up with that explains why the front end of the tug gets "lighter".

Francesca, the following is not a reply to your diagram with the red and blue arrows. I didn't see that until after I posted the following.

With no weight distribution applied:
1) the ball coupler pushes downward on the ball,
2) the TV's rear axle serves as a fulcrum, and
3) the TV's front axle gets "lighter" (load is removed).

If the distance from the rear axle to the ball is equal to one-half the distance from the rear axle to the front axle, the load removed from the front axle will be equal to one-half of the tongue weight.

With weight distribution applied, each lift chain might be tensioned to 1000#, giving a total tension of 2000# for two chains. This means a downward force of 2000# is applied to the TT's A-frame about 30" behind the ball coupler.
With the ball coupler acting as a fulcrum, and assuming the distance from coupler to TT's axles is 240", the 2000# downward force acting 30" behind the ball would be countered by an added force of 2000#*30"/240" = 250# acting upward against the TT's tires.

Now, the TV's rear axle becomes a fulcrum, and the added force acting upward against the TT's tires causes an added upward force to act against the TV's front tires
If the distance from the TV's rear axle to TT's axles is twice the distance form the TV's rear axle to front axle, the amount of load added to the front tires will be equal to twice the load added to the TT's tires.

In other words:
4) the tension in the WD bar lift chains exerts a downward force on the A-frame,
5) the downward force on the A-frame (with ball acting as fulcrum) causes an upward force on the TT's tires (load is added),
6) the upward force on the TT's tires (with TV's rear axle acting as fulcrum) causes an upward force on the TV's front tires (load is added), and
7) the upward force on the TT's tires (with TV's front axle acting as fulcrum) causes load to be removed from the TV's rear tires.

The load removed from the rear tires, via application of WD, will be equal to the sum of the loads added to the front tires and TT's tires.

Ron