WDH and airbag setup Question

[cruz-in]

So here is my setup:

2010 F250 Diesel 2WD CC Tow vehicle
11.5K trailer fully loaded
1200 to 1250 Pound Tongue weigh once loaded
Hensley Arrow hitch
1400 pound Springbars
Truck equiped with airbags

Been reading lots (and practcing) and pertty much got the setup up down but not sure about airing up the airbags. so far I have been:

1) Setting weight distribution so front bumper is about where it is whe not towing.

2) Then adjusting the airbags to level up the whole rig. Been taking 25 to 35 pounds of air in the bags.

Seems to drve real well except when I hit a bump there is a bit of porpoising. Nothing to dramatic, just a little bit annoying.

Is therre something I can do in my setup to limit the porpoising?

Thanks
Dan

[93Cobra2771]

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

[cruz-in]

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...

[Passin_Thru]

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

[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.

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

[2500HDee]

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.

[Ron_Gratz]

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

[Ron_Gratz]

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

[2500HDee]

...

[2500HDee]

Sorry, Ron corrected my incorrect FBD.

[2500HDee]

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.

[msmith1_wa]

I agree, raising the rear with air bags will not add weight back to the front.

[Ron_Gratz]

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

[lbrjet]

2500HDee wrote:
These discussions seem to be endless in forums across the internet. The weight distributing hitch is to restore the weight to the steer axle and put some of the tongue weight on the trailer axles.

The height of the rear of the tow vehicle does not matter and you can air up before or after you hook up your WDH but make sure you end up with the front at or slightly higher than the measured height before you connected the trailer.

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. This can be done the same way with a trailer on.

The easiest method is to:
1. Measure front ride height.
2. Connect trailer.
3. Adjust air bags.
4. Measure front ride height.
5. Adjust WDH to achieve front ride height at or slightly above step 1.

If you add too much air in step 3 then you won't be able to add any tension to the spring bars without bringing your front ride height below the initial level and therefore cannot transfer any weight to your trailer axles.

If the front is at or slightly (1/2" to an 1") above the initial value and IF the rear axle, tires, hitch, and payload of the truck can handle this amount of tongue weight, you don't even need a WDH. (This was the case with my last truck without even needing air bags.)

If not, you need to let out some air so that you can actually use the WDH.

This is about distributing the weight to the axles of the truck and trailer and is simple statics. This is not rocket science.

Requirements:
Front axle of truck -> should have the same or slightly less weight (height) than unloaded.

Rear axle of truck -> do not exceed RAWR (height does not matter)

Both axles of truck combined -> do not exceed GVWR

Hitch -> do not exceed weight rating (either with or without WDH, there are separate ratings)

Trailer axles -> do not exceed GVWR

Entire rig -> do not exceed GCVWR

How you arrive at the weights on each axle does not actually matter, only that you use what you have (WDH, airbags, self leveling suspension, unicorn farts) to keep the axle weights within these guidelines.

Excellent, excellent post.

[2500HDee]

These discussions seem to be endless in forums across the internet. The weight distributing hitch is to restore the weight to the steer axle and put some of the tongue weight on the trailer axles.

The height of the rear of the tow vehicle does not matter and you can air up before or after you hook up your WDH but make sure you end up with the front at or slightly higher than the measured height before you connected the trailer.

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. This can be done the same way with a trailer on.

The easiest method is to:
1. Measure front ride height.
2. Connect trailer.
3. Adjust air bags.
4. Measure front ride height.
5. Adjust WDH to achieve front ride height at or slightly above step 1.

If you add too much air in step 3 then you won't be able to add any tension to the spring bars without bringing your front ride height below the initial level and therefore cannot transfer any weight to your trailer axles.

If the front is at or slightly (1/2" to an 1") above the initial value and IF the rear axle, tires, hitch, and payload of the truck can handle this amount of tongue weight, you don't even need a WDH. (This was the case with my last truck without even needing air bags.)

If not, you need to let out some air so that you can actually use the WDH.

This is about distributing the weight to the axles of the truck and trailer and is simple statics. This is not rocket science.

Requirements:
Front axle of truck -> should have the same or slightly less weight (height) than unloaded.

Rear axle of truck -> do not exceed RAWR (height does not matter)

Both axles of truck combined -> do not exceed GVWR

Hitch -> do not exceed weight rating (either with or without WDH, there are separate ratings)

Trailer axles -> do not exceed GVWR

Entire rig -> do not exceed GCVWR

How you arrive at the weights on each axle does not actually matter, only that you use what you have (WDH, airbags, self leveling suspension, unicorn farts) to keep the axle weights within these guidelines.