WDH and airbag setup Question

With your LT rated tires, and the tongue weight that you have, I would be running max sidewall pressure...

Passin Thru wrote:
Maximum air P to tires. Air bag to 45 lb and try. Do this before you level it with the hitch.

Interesting my experimenting while towing yesterday indicates 35 to 45 pounds seems to geve the best ride...

Maximum air P to tires. Air bag to 45 lb and try. Do this before you level it with the hitch.

Ron, you are right. Thank you for the correction. It was the center of gravity's distance from the fulcrum that I was thinking of as you pointed out in the wheelbarrow post. I am going to delete my other posts as if to not leave incorrect information posted.

I am going to run some numbers on the COG of the truck and load in the bed though, as I do think that it is significant in the weight distribution of the truck. I am referencing a specific time that a kid I was roofing with loaded an entire pallet of shingles in an F150 and the front suspension was at full droop. And yes, he did have them loaded over the axle, not behind it.

Also, when setting up race cars we used 4 wheel scales and adjusted the coil overs up or down to balance the load side to side and or front to rear but I realize I am providing personal examples here and not numbers so I will get back with everyone soon.

Thanks for the discussion.

I just remembered, you can only do this for cross-corner weights, not for side to side or front to rear. So disregard that.

Ron Gratz wrote:
Let's say the length of your beam is "WB". Since it's a uniform beam, the distance from the front axle to the CG is WB/2 and the distance from the CG to rear axle is WB/2. With the beam at an angle, "theta", the horizontal distances from front axle to CG and from CG to rear axle are cosine(theta)*WB/2.

To maintain rotational equilibrium, the sum of the moments about any point must equal zero. For your example below, the 25# force at the front generates a CW moment equal to 25*cosine(45 degrees)*WB/2. The 75# force at the rear generates a CW moment equal to 75*cosine(45 degrees)*WB/2. Clearly, your free body cannot be in rotational equilibrium unless the upward force at the front and the upward force at the rear are equal. The end reactions shown in the diagram below must both be equal to 50#.

In general the sum of moments about the CG must give:
Ff*cosine(theta)*D1 - Fr*cosine(theta)*D2 = 0
where Ff and Fr are the vertical upward forces acting at front and rear, D1 is distance from front axle to GC, and D2 is distance from CG to rear axle.
Therefore, we have: Ff*D1 = Fr*D2
And we know: Ff+Fr = W = weight of beam
These two equations give: Ff = W*D2/(D1+D2) and Fr = W*D1/(D1+D2)

The distribution of weight is not dependent on the value of "theta".

When you stand the beam vertically on end, as you did in your first diagram, the front reaction force, the rear reaction force, and the weight force are all collinear. The values of D1, D2, and D1+D2 are all zero. There are no moments acting about the CG. You have a case of unstable equilibrium. Ff and Fr could be any value as long as their sum is equal to the weight. However, in the limiting process of approaching 90 degree slope, we know that the above equations for Ff and Fr apply.

Ron

Following diagram provided by 2500HDee

Ron, you are right. Thank you for the correction. It was the center of gravity's distance from the fulcrum that I was thinking of as you pointed out in the wheelbarrow post. I am going to delete my other posts as if to not leave incorrect information posted.

I am going to run some numbers on the COG of the truck and load in the bed though, as I do think that it is significant in the weight distribution of the truck. I am referencing a specific time that a kid I was roofing with loaded an entire pallet of shingles in an F150 and the front suspension was at full droop. And yes, he did have them loaded over the axle, not behind it.

Also, when setting up race cars we used 4 wheel scales and adjusted the coil overs up or down to balance the load side to side and or front to rear but I realize I am providing personal examples here and not numbers so I will get back with everyone soon.

Thanks for the discussion.

2500HDee wrote:
Ron, the amount of weight is proportional the the ride height of the front. If the ride height is back to where you started, 100% of the weight is restored. You can achieve this by raising the rear even higher than you started if needed although that increases caster and makes the truck steer funny.

The amount of load removed from the front axle is directly proportional to the vertical downward force on the ball, directly proportional to the distance from rear axle to ball, and inversely proportional to the TV's wheelbase. The amount of front end rise is directly proportional to the amount of load removed and inversely proportional to the spring rate. You cannot change the front-end load by raising the rear -- unless you have a trailer attached with WD applied.

Think about a wheelbarrow, the higher you lift the handles, the more weight goes on the front wheel and the lighter the handles get. When you dump it you stand it right on the wheel and the handles are weightless.

The CG of the load in a wheelbarrow is considerably higher than the axle and relatively close to the axle. As you lift the handles, the horizontal distance from load CG to axle decreases quickly making it easier to raise the handles.
Raising the rear end of a TV is nothing like lifting the handles of a wheelbarrow. There is virtually no change in horizontal distance from CG to pivot point. If you get the CG over the front pivot -- you're in big trouble.

Ron

Let's say the length of your beam is "WB". Since it's a uniform beam, the distance from the front axle to the CG is WB/2 and the distance from the CG to rear axle is WB/2. With the beam at an angle, "theta", the horizontal distances from front axle to CG and from CG to rear axle are cosine(theta)*WB/2.

To maintain rotational equilibrium, the sum of the moments about any point must equal zero. For your example below, the 25# force at the front generates a CW moment equal to 25*cosine(45 degrees)*WB/2. The 75# force at the rear generates a CW moment equal to 75*cosine(45 degrees)*WB/2. Clearly, your free body cannot be in rotational equilibrium unless the upward force at the front and the upward force at the rear are equal. The end reactions shown in the diagram below must both be equal to 50#.

In general the sum of moments about the CG must give:
Ff*cosine(theta)*D1 - Fr*cosine(theta)*D2 = 0
where Ff and Fr are the vertical upward forces acting at front and rear, D1 is distance from front axle to GC, and D2 is distance from CG to rear axle.
Therefore, we have: Ff*D1 = Fr*D2
And we know: Ff+Fr = W = weight of beam
These two equations give: Ff = W*D2/(D1+D2) and Fr = W*D1/(D1+D2)

The distribution of weight is not dependent on the value of "theta".

When you stand the beam vertically on end, as you did in your first diagram, the front reaction force, the rear reaction force, and the weight force are all collinear. The values of D1, D2, and D1+D2 are all zero. There are no moments acting about the CG. You have a case of unstable equilibrium. Ff and Fr could be any value as long as their sum is equal to the weight. However, in the limiting process of approaching 90 degree slope, we know that the above equations for Ff and Fr apply.

Ron

Following diagram provided by 2500HDee

...

Sorry, Ron corrected my incorrect FBD.

Ron Gratz wrote:

2500HDee wrote:
Both the WDH and airbags/stiffer leafs/add-a-leafs/Timbrens/etc will affect the weight on the front axle. The difference is the rear suspension will change this regardless of whether a trailer is attached or not. Ever seen that guy with a whole pallet of shingles in his F-150? The front tires almost come off the ground on every bump. Since he can't use a WDH to transfer the weight to the front axle of his truck full of shingles, the other option, assuming it would not exceed the RAWR or GVWR of his truck, is to have stiffer rear springs. By bringing the back of the truck back up it will restore weight to the front axle and make the truck safe to drive. This can be illustrated with a simple free body diagram.---

Can you post (or describe with words) a simple free body diagram which will show how using air bars to bring the back of the truck up will restore load to the front axle?

Also, can you provide an estimate of how much load is removed from the front axle for a given weight of shingles, and how much load is restored to the front axle by raising the rear with air bags?

Finally, do you have any scales measurements to substantiate this claim?

Ron

Sure thing, I will get back to you after work tonight.