Weight distributing hitch math?

Ron Gratz wrote:
When you drop the coupler on the ball without using WD,
a load equal to about 40-50% of the tongue weight will be removed from the front axle, and
a load equal to about 140-150% of TW will be added to the rear axle.
The combined added load on the TV's front and rear axles will be equal to 100% of the tongue weight.

If you then adjust the WDH to restore all of the load removed from the TV's front axle,
a load equal to about 20-25% of TW will be transferred to the TT's axles.
A load equal to about 40-50% of TW will be added to the TV's front axle, and
a load equal to about 60-75% of TW will be removed from the TV's rear axle.

The net load change on the front axle due to TW and WD will be zero, and
the net load change on the rear axle will be an addition equal to about 75-80% of TW.
The net load change on the TV's axles will be an addition equal to about 75-80% of TW.

So, you are correct -- with a TW of 500#,
when the WDH is adjusted to transfer a load equal to 20-25% of TW to the TT's axles,
the load on the TT's axles will increase by about 100-125#, and
the load on the TV's axles will decrease by about 100-125#.

In compliance with the law of conservation of mass,
mass has been neither created nor destroyed by the WD process.

Ron

I'm struggling to fully grasp the concepts here, so please bear with me if this is a dumb question:

Is the "force" required to shift the weight measurable, and if so how much is required in the above example?

Also, where is that force exerted/directed???