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Old 11-02-2015, 09:22 PM   #81
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I know your focus is on tire overloading but I took time to check the gross on my single axle trailer, it is 3510 lbs. and the rims are rated at 1760 lbs. so there is only 20 lbs. more rim capacity than the gross of the trailer. Side loading of the tire is also side loading the rim and if the rim fails the result IMHO is worse than the tire failing though neither one is a picnic.

Really seems to me the manufacturers should be REQUIRED to install equipment with more capacity.
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Old 11-02-2015, 10:08 PM   #82
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I know your focus is on tire overloading but I took time to check the gross on my single axle trailer, it is 3510 lbs. and the rims are rated at 1760 lbs. so there is only 20 lbs. more rim capacity than the gross of the trailer. Side loading of the tire is also side loading the rim and if the rim fails the result IMHO is worse than the tire failing though neither one is a picnic.

Really seems to me the manufacturers should be REQUIRED to install equipment with more capacity.
IMO RV companies for the most part appear to have made the decision to cut corners and to only do the minimum required by federal regulation rather than spend a few hundred and deliver a much better, and safer vehicle to their customers. The requirement is that the GAWR be less than the total load capacity of the tires (and wheels). They have chosen to ignore the fact that side to side and axle to axle loads are not evenly split so they must, or should know that with load margins less than 50# in some cases one tire is almost guaranteed to be in constant overload even before considering side loading to curves or wind.
I will leave it up to you to decide if companies that cut things that close really are building the "quality RVs" they claim.
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Old 11-02-2015, 10:23 PM   #83
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I've changed focus from overloading on curves to looking at tire overloading due to cross wind. I've now learned how to calculate wind force using, among other things, an SAE technical document that analyzes tractor trailer instability in wind and on ice as a model.

An important question is how much wind is reasonable to assume under normal conditions. California and Oregon both issue wind speed advisories when the sustained wind is over 35mph. Other agencies seem to use different thresholds, some of which are higher. It seems reasonable for the moment to assume that if a wind advisory hasn't been issued, conditions are 'normal'.


If my math is right, when pulling my fully loaded trailer in a crosswind of 35mph, the leeward tires will be overloaded by 425lbs each. This could go on for a long time without my knowledge.


If my tires had the 20% margin recommended by Carlisle Tire, there would be no tire overload with a 35mph cross wind.


This math needs checking.
I'll provide my spreadsheet to anyone who's willing to check it.
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Old 11-04-2015, 12:34 AM   #84
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I'm horrible with the numbers but, wind loading on rigging for overhead lifting is huge. Depending on surface area, you can see huge jumps in actual weight/ load cell.
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Old 11-04-2015, 10:38 AM   #85
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For anyone who might be willing to check it, or wants to try using it, here’s my math for calculating the increased load on leeward tires due to a cross wind. Don't glaze over. The math turns out to be just multiplies and divides. The numbers I use are for a 35mph wind and the dimensions of my trailer. You should be able to copy what I did using numbers for your own choice of wind speed and trailer.


1. Calculate the force of wind on the side of the trailer

A travel trailer has a lot in common with a tractor trailer.The following equation comes from “Wind Effects On Dynamic Stability of Tractor Trailers In Winter Conditions”, published by the SAE.

Fwind = 0.5 * Rho * Cd * Vwind * Vwind * L * H

Rho = 0.002376
This is the density of air and has the units of slug per cubic foot.
I had to look up ‘slug’.It’s mass in English units using feet and lbs for distance and force.

Cd = 2.0
This is the ‘Drag Coefficient’ that depends on the shape of the object.
The value of 2.0 is used for tractor trailers by the SAE article.

Vwind
This is the speed of the wind in units of ft/sec.
To get ft/sec, multiply mph by 5280/3600.
At 35mph, the wind speed is 51.33 ft/sec.

L and H
These are the length and height of the trailer box in units of feet.
Notice that they’re multiplied together in the equation to produce area.
For my trailer, I’m using L = 26ft and H=8ft.

So,Fwind = 0.5 * 0.002376 * 2.0 * 51.33 * 51.33 * 26 * 8=1302lbs



2. Calculate the tipping torque on the trailer

The force of the wind pushing on the side of the trailer creates a torque or moment that tries to tip the trailer over.The force from the wind is equivalent to a single force pressing in the middle of the side of the trailer. The higher the trailer, the more the torque.Multiplying total force times the height above the road surface of the center of the trailer side wall gives tipping torque.I’m using 5ft as the center of my wall.

Tipping Torque = Fwind * CenterHeight= 1302 * 5 = 6510 ft-lbs



3. Calculate the restoring force on the tires

The trailer doesn’t tip over because the forces on the tires create a restoring torque that keeps the trailer upright.Force transfers from the windward tire to the leeward tire.This works until the windspeed is so large that the force on the windward tire is reduced to zero, at which point the trailer rolls over.Up to rollover, the increased force on the leeward tires, and reduced force on the windward tires creates a torque that balances the tipping torque.The wider the trailer, and further spaced the wheels, the less force it takes to do this.Like the SAE article, I use the leeward tires as the pivot point.So, to calculate restoring force from tipping torque, I divide by the distance between wheels, which in my case is about 7.5ft.


Restoring Force = Tipping Torque / Wheel Span = 6510 / 7.5 = 868 lbs

4. Calculate the force due to wind per tire

The total Restoring Force is spread among all the tires on one side.Divide by the number of axles to get the added load per tire due to wind.

Tire Wind Load = Restoring Force / #Axles = 868 / 2 = 434 lbs per tire


So, if my math is right, a 35mph wind will cause an increase in load of 434 lbs on each of my downwind tires. This is approximate because the drag coefficient in the original force equation is approximate.


If someone finds a fundamental error in this, I'll redo it.
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Old 11-04-2015, 10:41 AM   #86
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If someone finds a fundamental error in this, I'll redo it.
You forgot to factor in the rotational mass of the schwansoculator.
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Old 11-04-2015, 10:44 AM   #87
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You forgot to factor in the rotational mass of the schwansoculator.
I think you mean 'framistat', because it's what gets rectabularly extruded, not the schwansoculator.
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Old 11-04-2015, 11:50 AM   #88
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I think you mean 'framistat', because it's what gets rectabularly extruded, not the schwansoculator.
I thought it was the dual focal metering valve.

FWIW, Gyrogearloose, I haven't run that formula against my TT but those load numbers you have look about right to me.
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Old 11-04-2015, 11:56 AM   #89
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I thought it was the dual focal metering valve.

FWIW, Gyrogearloose, I haven't run that formula against my TT but those load numbers you have look about right to me.
And, of course, the implication is that if there's no margin between the load on my tires and their load rating when the trailer is sitting still in my driveway, then the tires are overloaded by hundreds of pounds when I encounter a significant crosswind (but not strong enough to trigger a high wind advisory).
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Old 11-04-2015, 12:03 PM   #90
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So, given all that what is the solution? If the senario of it that being already overloaded or near over loaded before moving and given the force of rotation on curves and coupled in with wind velocity then what are the options? Take curves at a low speed, pullover and stop during wind conditions, switch out axles, tires and rims to larger, higher capacity?
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