Dual Cam Setup

JBarca wrote:

We need mrekim to confirm his loaded TW and how he had the WD adjusted at the time the snap up and frame damage occurred. I was under the impression that he had an 1,100# loaded TW and had adjusted the WD hitch to return the front axle very close to unhitched fender height. I may have mixed that up, we need him to confirm. If all his damage came from only a 500# per side load down onto the WD bar, that is not a comforting thought.

My scale TW, which should be close to what I was when the damage occurred (could be less, but not more), was 1080.

JBarca wrote:

Ron Gratz wrote:

In this post, I attempted to address the increase in WD bar load for a sway angle of 5 degrees.

I cannot get the link to fire. Is the link OK or it my PC?

It doesn't work for me either.

Ron,

Thank you for breaking down the detail response. Very much appreciated. Some comments inserted.

Ron Gratz wrote:

JBarca wrote:
---I'm assuming you just picked the 500# as a number to use in the calculation not knowing actual loaded TW's and TV/TT stats. If this correct?

The 500# per bar is a rough approximation based on the OP's scales data which showed 220# returned to the front axle when WD was applied.
If I had used a value of 1000# per bar, the calculated values would have been doubled.

We need mrekim to confirm his loaded TW and how he had the WD adjusted at the time the snap up and frame damage occurred. I was under the impression that he had an 1,100# loaded TW and had adjusted the WD hitch to return the front axle very close to unhitched fender height. I may have mixed that up, we need him to confirm. If all his damage came from only a 500# per side load down onto the WD bar, that is not a comforting thought.

Ron Gratz wrote:
For purposes of sway control, the longitudinal tension/compression in the WD bar acts like the tension/compression in a friction sway bar -- it resists a change in the angle between TV and TT.
It is interesting to note that a friction sway bar is factory-set to produce a friction force of 1100# compared to the WD bar tension of 2000# which I calculated for the 500# bar load.

You may have just confirmed with calculations why a single friction bar is not effective on larger/longer trailers and the DC system is. And you were only dealing with the rear side of the DC, not adding front side friction on the opposite side the trailer. It may also help add to why some ORF members have reported their rig had acceptable results on longer trailers when using 2 friction sway bars.

Ron Gratz wrote:

JBarca wrote:

---The large tension on the right side of the A frame appears to be rolling the shank and possibly the truck a little. The shank for sure has a CW tilt as viewed from the TT. With your recent found force magnification in this extreme turn, this possibly might add new light to making a turn with the DC and the large loads that can come from it.

When there is a large TV/TT angle, any upward force on the rear end of a WD bar will tend to rotate the TV about its roll axis and about its pitch axis.
Because the TV's track width is much less than its wheelbase, it is much easier to make it rotate about the roll axis.
When the hitch head is able to rotate CW as shown in your photo, the slope of the front of the WD bar is reduced causing the load on the bar to be reduced.

JBarca wrote:
---If the inside WD bar is unloaded by some amount, then WD is potentially reduced on left side bar leaving the right side WD to attempt/or add to carrying more of the WD load.

In a sharp turn, the slope of the front end of the inside WD bar will be significantly reduced and the load on that bar can go to zero.
However, that does not mean the load removed from the inside bar will be added to the outside bar.
As your photo shows, on the level surface, there was a left to right tilt of the hitch head because it is relatively easy to rotate the TV about its roll axis.
This means the outside WD bar might actually be subjected to less load than when the TV/TT angle was zero.

JBarca wrote:
---Does the roll action then even more increase the chain tension and further increase the friction force? Granted this turn is extreme in this case, however a 50 degree turn with a compound angle of 5 to 8 degrees of the hitch head can unload to zero the inside turn WD bar which can be assumed did happen in mrekins case.

I agree the inside bar can unload to zero, but that is a result of the decreasing tilt of the trunnion axis as the hitch head swings relative to the trailer.

From mrekim’s thread and other recent hitch head and trunnion bar interaction investigation, I no longer can be sure that the WD chain tension will increase as the WD bar rides up the cam pending what the hitch head is doing. There is a hitch head component to this that is still in the discovery stage.

However I still have a “belief” that in a compound angle turn when the inside turn WD bar can unload to zero or closer to zero, there is “some” level of increased tension in the outside turn WD bar. How much tension increase is still unknown. Here it the thought path (may be flawed, but this is how I came to it).

When the TV and TT are on level ground and straight ahead, both WD bars have very close to equal loads. Those WD loads came from the setup of the WD hitch. As a result of this, both left and right WD chain loads are close to the same and the trailer A frame is close to being equally loaded both left and right. So far so good and I think we all agree on this.

Now we come to the point where some DC owners have had snap up failures and in mrekims’ case, A frame failures. A turn has been associated with many if not all snap up failures and even Reese tech service told me in some cases, the snapup can fail with heavy TW’s if the snap up is not bolted. (Note: The Reese tech call was before the newer HD snapup was on the market, the comment relates to the non HD snap up)

Now enter the compound angle turn. When the hitch head rolled and unloaded the inside turn WD bar to zero or near zero, there is only the outside turn bar left to provide any level of WD. If the outside bar load stayed the same, then the truck would of lost 1/2 the WD and the front of the truck would of came up enough the driver “may” feel it. Lighter suspension trucks would fell it more so then stiffer suspension trucks.

I do compound angle turns all the time, and in none of my trucks, 1/2, 3/4 or 1 ton I do not recall the front end going extremely light. We know the WD bar is riding up the cam and we know the rear of the WD bar is rising. We know the trunnion lug on that side is at a down hill angle due to the roll axis. We do not know what the trunnion lug angle in the hitch head did to this equation. That is still an open topic.

Since the TT TW did not change, it is still providing a constant force down on the only WD bar that can produce tension. Since the same TW mass is available to only 1 WD bar it comes down to, will the hitch head angle and the rising rear of the WD bar allow an increase in tension?

While the trunnion lug angle in the hitch head may have relieved some of the WD tension, more of the existing TW is available to be held by the outside turn WD bar. An unknown is, did the hitch head angle change enough to not allow more WD bar tension on the outside turn bar and did the rear of the bar riding up the cam over come the loss in the hitch head?

In order for the snap up to fail like they do, an increased force presents itself to one side of the WD hitch at the WD chain that overloads a non bolted snapup. A hypothesis is, that the outside WD bar rose enough in tension from the compound angle turn to cause the snap up failure. Now the task is, how to quantify that? If this compound angle turn is not creating the force, then what other combination of actions is?

Ron Gratz wrote:

In this post, I attempted to address the increase in WD bar load for a sway angle of 5 degrees.

I cannot get the link to fire. Is the link OK or it my PC?

Thanks

John

mrekim wrote:
I'm trying to make sure I understand. Your increase from 1000# to about 1075 above is just the chain tension force due the the bar being tensioned more due to an additional 3/4" distance. This part I understand, it's just like tightening up the chain by 1/4 link.

Yes, that's correct -- except my example had the bar being lifted by 1/4" rather than 3/4".

Then when sway occurs, there's some additional (large?) friction force that occurs as the bar tries to climb the 45 degree bend at the end of the bar. This force only occurs during motion (sway or turning) Is that correct?

Friction force occurs whenever there is a force on the bar which tries to make the bar move relative to the cam.
Since sway or turning with a varying turn radius will make the bar move relative to the cam, the friction force would occur during those actions.
The cam/bar friction force acts to resist any change in the TV/TT angle.

If yes, then during sway does this friction occur on both sides since for a small angle the spring bar is climbing the cam on both sides?

Let's assume the TV/TT angle is zero and both cams are perfectly centered. Each cam will be in contact with both the front slope and the rear slope of its bar.

As soon as the TV/TT angle begins to change from zero,
the bar on the inside of the bend loses contact with the rear slope of its bar and the front slope begins to slide rearward and up over its cam.
The bar on the outside of the bend loses contact with the front slope of its bar and the rear slope begins to slide forward and up over its cam.

So, the short answer to your question is -- for a small angle of increasing magnitude, the both WD bars are sliding up their respective cams, and both bars are generating friction force.

Both bars will be resisting the swing away from the "centered" position.
Since the magnitude of a bar's rear slope usually is greater than the magnitude of the front slope, the outside bar usually will generate significantly more "sway resistance" than will the inside bar.

Ron

Highway 4x4 wrote:
Wow, I am sure the DC system works as many just love it, but trying to understand how is making my head hurt. It looks kind of complicated and prone to issues of many types. I am not saying it's not good but 17 pages is really not helping to sell it. For those reasons, I'm out.

Just because there are a lot replies does not make it complicated to install or use. No need to be "out". After all, there are the exact same number of replied to this thread, which is even more complicated and possibly hard to understand, yet just about everybody uses one, or should, if they have a large trailer. :)
Barney

Ron Gratz wrote:

For a large sway angle of, say, 5 degrees, a cam will move about 0.5" longitudinally and, due to the slope of the bar, will lift the bar about 0.25". The added upward force on the bar will be equal to 0.25" times the spring constant. For an 800# bar with a base load of 1000#, the force will increase by 0.25"x200#/inch = 50#. The upward force will increase from 1000# to about 1050#. For a 1200# bar with a base load of 1000#, the force will increase from 1000# to about 1075#.

IMO, the "claim to fame" for the Dual Cam is the fact that the cam/bar friction force is much greater when the TT is trying to swing away from centered than it is when the TT is trying to swing toward centered.

I'm trying to make sure I understand. Your increase from 1000# to about 1075 above is just the chain tension force due the the bar being tensioned more due to an additional 1/4" (fixed typo) distance. This part I understand, it's just like tightening up the chain by 1/4" (fixed typo).

Then when sway occurs, there's some additional (large?) friction force that occurs as the bar tries to climb the 45 degree bend at the end of the bar. This force only occurs during motion (sway or turning) Is that correct?


If yes, then during sway does this friction occur on both sides since for a small angle the spring bar is climbing the cam on both sides?

Wow, I am sure the DC system works as many just love it, but trying to understand how is making my head hurt. It looks kind of complicated and prone to issues of many types. I am not saying it's not good but 17 pages is really not helping to sell it. For those reasons, I'm out.

JBarca wrote:
After thinking about this, the friction load cannot be linear with the DC once the WD keeps moving. One of the underlying claims to fame which makes this anti sway control so effective is the WD bar increases in tension as the WD rides up the cam. The increased tension increases the friction holding action.

I've seen that claim to fame -- but I've never seen any substantiation of it.

In this post, I attempted to address the increase in WD bar load for a sway angle of 5 degrees.
My analysis gave the folloing:
For a large sway angle of, say, 5 degrees, a cam will move about 0.5" longitudinally and, due to the slope of the bar, will lift the bar about 0.25". The added upward force on the bar will be equal to 0.25" times the spring constant. For an 800# bar with a base load of 1000#, the force will increase by 0.25"x200#/inch = 50#. The upward force will increase from 1000# to about 1050#. For a 1200# bar with a base load of 1000#, the force will increase from 1000# to about 1075#.

IMO, the "claim to fame" for the Dual Cam is the fact that the cam/bar friction force is much greater when the TT is trying to swing away from centered than it is when the TT is trying to swing toward centered.

Now relearning that... :S what assumptions did you use to create the 2,170# breakaway force? Is that the initial breakaway friction force pulling the WD bar up hill? Meaning only at one point in time as the WD bar 1st starts to slip on the cam. As once the bar starts sliding, the friction lowers slightly as it is sliding friction vrs static however the WD force is believed to be increasing riding up the cam which will increase the friction force nce again.

What did you use for the coefficient of friction? 0.5 to 0.7?

As stated on my diagram, the friction force is assumed equal to 0.6 times the normal force. That means the assumed coefficient of friction is 0.6. Since the slip speed between bar and cam is quite small, there probably is very little difference between the static coefficient of friction and the dynamic coefficient.
Again, if we're talking about sway conditions as opposed to sharp turns, there is very little relative movement between bar and cam and very little change in WD bar load.

Ron

JBarca wrote:
---I'm assuming you just picked the 500# as a number to use in the calculation not knowing actual loaded TW's and TV/TT stats. If this correct?

The 500# per bar is a rough approximation based on the OP's scales data which showed 220# returned to the front axle when WD was applied.
If I had used a value of 1000# per bar, the calculated values would have been doubled.

On Edit: The 200# value was from a preliminary weighing.
Subsequent re-adjustment of the OP's WDH caused the TV's front axle to be returned approximately to the unhitched height.
If we assume the WDH caused 560# to be returned to the front axle,
the corresponding WD bar load would be about 1000# per bar.

For purposes of sway control, the longitudinal tension/compression in the WD bar acts like the tension/compression in a friction sway bar -- it resists a change in the angle between TV and TT.
It is interesting to note that a friction sway bar is factory-set to produce a friction force of 1100# compared to the WD bar tension of 2000# which I calculated for the 500# bar load.

My analysis does not depend on any assumptions about TV/TT angle or left to right tilt of the hitch head.
It only assumes the WD bar produces a given amount of vertical force against the cam and assumes enough change in TV/TT angle to make the end of the WD bar slide up and forward along the back side of the cam.
The effect of the friction will be present when swinging from zero to five degrees just as it would be when swinging from 40 to 45 degrees.

---The large tension on the right side of the A frame appears to be rolling the shank and possibly the truck a little. The shank for sure has a CW tilt as viewed from the TT. With your recent found force magnification in this extreme turn, this possibly might add new light to making a turn with the DC and the large loads that can come from it.

When there is a large TV/TT angle, any upward force on the rear end of a WD bar will tend to rotate the TV about its roll axis and about its pitch axis.
Because the TV's track width is much less than its wheelbase, it is much easier to make it rotate about the roll axis.
When the hitch head is able to rotate CW as shown in your photo, the slope of the front of the WD bar is reduced causing the load on the bar to be reduced.

---If the inside WD bar is unloaded by some amount, then WD is potentially reduced on left side bar leaving the right side WD to attempt/or add to carrying more of the WD load.

In a sharp turn, the slope of the front end of the inside WD bar will be significantly reduced and the load on that bar can go to zero.
However, that does not mean the load removed from the inside bar will be added to the outside bar.
As your photo shows, on the level surface, there was a left to right tilt of the hitch head because it is relatively easy to rotate the TV about its roll axis.
This means the outside WD bar might actually be subjected to less load than when the TV/TT angle was zero.

A question is, is this then a self multiplying force situation? First the DC friction effect creates a high load on one side of the A frame by increased chain tension. This increased tension can then roll the hitch head by some amount.---

No, it does not. The increase in chain tension, in my analysis, results from the compression in the cam bar which is due to the increased normal and friction force acting on the cam as the WD bar tries to slide up and over the cam.
The increased chain tension pulls down on the A-frame, but the compression in the cam bar exerts an equal upward force on the A-frame. The net downward force acting on the A-frame remains at, for my example, 500#.

---Does the roll action then even more increase the chain tension and further increase the friction force? Granted this turn is extreme in this case, however a 50 degree turn with a compound angle of 5 to 8 degrees of the hitch head can unload to zero the inside turn WD bar which can be assumed did happen in mrekins case.

I agree the inside bar can unload to zero, but that is a result of the decreasing tilt of the trunnion axis as the hitch head swings relative to the trailer.

Using those spec's I'm assuming you can calculate the tension in the WD chains when on level ground and the TV & TT are in a straight line. That puts this at a starting place of chain load when going straight ahead.---

Assuming zero pitch-axis rotational torque between ball and coupler:
The load removed from the front axle would be 1260*(65/130) = 630#.
To restore 630# to the front axle, the combined WD bar force would have to be 630*130/(65+230+16) = 263#.
Assuming longitudinal distance from ball to snap-up brackets is 30", the combined chain tension would have to be 263*(230+16)/30 = 2157# or approximately 1080# per chain.
Of course, that assumes travelling straight ahead so there is no longitudinal tension/compression in the WD bars -- i.e. no friction force between WD bar and cam.

Since we do not yet know an accurate spring constant for the WD bar it may be a little hard to know the full tension in the chain when in the turn and what effect the hitch head roll created. The assumption is, the chain force will even be higher then the straight ahead chain load.

I would not assume the force will be even higher. It depends on how much rearward tilt there is when the TV/TT angle is zero and how much the rearward tilts of the trunnion axes change as the TV/TT angle changes.

Another question/topic: Did I understand your last statement correctly. For a 500# downward starting force from the TT acting on one WD bar, this translates into a 2,170# at the WD chain due to the increase from the breakaway friction load of the turn? And if so, is this then a 2,170/500 = 4.3 to 1 linear increase in WD chain load per pounds of down force? If linear, is it then if we have 1,000# downward force (fill in the number as needed) from the TT do we end up with a 4,300# WD chain load in a turn from the breakaway friction force riding up the cam?

I addressed this earlier. If my analysis is correct, a load of 1000# per WD bar and the other assumptions given in my free body diagram presentation would give a chain tension of about 4300# per chain.
Keep in mind that this is a dynamic process -- the TV/TT angle must be changing in order to generate the friction force.
If the TV and TT were being driven in a constant-radius circle, the TV/TT angle would not be changing and there would be no friction force.

Ron

JBarca wrote:
Ron,

Another question/topic: Did I understand your last statement correctly. For a 500# downward starting force from the TT acting on one WD bar, this translates into a 2,170# at the WD chain due to the increase from the breakaway friction load of the turn? And if so, is this then a 2,170/500 = 4.3 to 1 linear increase in WD chain load per pounds of down force? If linear, is it then if we have 1,000# downward force (fill in the number as needed) from the TT do we end up with a 4,300# WD chain load in a turn from the breakaway friction force riding up the cam?

After thinking about this, the friction load cannot be linear with the DC once the WD keeps moving. One of the underlying claims to fame which makes this anti sway control so effective is the WD bar increases in tension as the WD rides up the cam. The increased tension increases the friction holding action.

Now relearning that... :S what assumptions did you use to create the 2,170# breakaway force? Is that the initial breakaway friction force pulling the WD bar up hill? Meaning only at one point in time as the WD bar 1st starts to slip on the cam. As once the bar starts sliding, the friction lowers slightly as it is sliding friction vrs static however the WD force is believed to be increasing riding up the cam which will increase the friction force once again.

What did you use for the coefficient of friction? 0.5 to 0.7? Conditions may not be perfect in the this setting.

Ron,

Great post! And yes, amazing to see the magnification up of the 500# force. I'm assuming you just picked the 500# as a number to use in the calculation not knowing actual loaded TW's and TV/TT stats. If this correct?

Let me add some to that picture and maybe you can carry this to the next level. The TT picture is from my current camper when I had my K2500 Suburban. I will list setup parameters when that picture was taken. A few points to note, as you can see (concrete speckles under the camper) I was turning around on my concrete pad in my drive way. The concrete was level however the hitch head had some left to right tilt. That turn is the physical mechanical limit of the Suburban and hitch on a hard left turn. It was approx 74 degree left. If I went any further I would of broke the hitch.

Something I now see since that pic was taken, and may add to this topic of increasing chain forces on one side of the A frame even on level ground. You can see the hitch shank it tilted when the truck was on level concrete. The large tension on the right side of the A frame appears to be rolling the shank and possibly the truck a little. The shank for sure has a CW tilt as viewed from the TT. With your recent found force magnification in this extreme turn, this possibly might add new light to making a turn with the DC and the large loads that can come from it.

If the heavy down force on the right side of the A frame rolled the hitch head it could have the ability to unload some of the inside WD bar due to the head tilt. By how much, I really do not know in this case. This is not a compound angle turn between TV and TT yet the large turn and the heavy down force you show is large enough to rotate the hitch head. If the inside WD bar is unloaded by some amount, then WD is potentially reduced on left side bar leaving the right side WD to attempt/or add to carrying more of the WD load.

A question is, is this then a self multiplying force situation? First the DC friction effect creates a high load on one side of the A frame by increased chain tension. This increased tension can then roll the hitch head by some amount. Does the roll action then even more increase the chain tension and further increase the friction force? Granted this turn is extreme in this case, however a 50 degree turn with a compound angle of 5 to 8 degrees of the hitch head can unload to zero the inside turn WD bar which can be assumed did happen in mrekins case.

Here is larger pic of the one you posted. Look at the shank angle.


Here is the inside turn bar


Some stats of that setup so we can see how this multiplies up. (I actual have scaled axles weights of that setup if needed.)

Loaded camper TW: 1,260#
WD bars size: 1,200#

Suburban WB: 130"
Suburban Rear Overhang: 65"
WD was set return the Suburban to unhitched front axle weight
Here are the TT dimensions


Using those spec's I'm assuming you can calculate the tension in the WD chains when on level ground and the TV & TT are in a straight line. That puts this at a starting place of chain load when going straight ahead. Since we do not yet know an accurate spring constant for the WD bar it may be a little hard to know the full tension in the chain when in the turn and what effect the hitch head roll created. The assumption is, the chain force will even be higher then the straight ahead chain load.

Another question/topic: Did I understand your last statement correctly. For a 500# downward starting force from the TT acting on one WD bar, this translates into a 2,170# at the WD chain due to the increase from the breakaway friction load of the turn? And if so, is this then a 2,170/500 = 4.3 to 1 linear increase in WD chain load per pounds of down force? If linear, is it then if we have 1,000# downward force (fill in the number as needed) from the TT do we end up with a 4,300# WD chain load in a turn from the breakaway friction force riding up the cam?

Thanks

John