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Open rear diff. What a freakin' joke

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Check your figures

A 100% grade is 90 Degrees. And a 21% grade does calc to 18. 9 degrees.

I am a surveyor so trust me. If not go to aj.com and use a grade to degree calculator:confused: :confused:
 
Here is what I found at aj.com:



Question



I live in the mountains and roads are always rated by their "percent grade". What does percent grade mean, and how is it related to degrees?



Answer







When you travel through the mountains, it is very common to see signs that say things like, "Trucks check brakes -- 10% grade" or "6% grade -- Trucks use right lane only".

The percent grade is a simple calculation. Let's say you have a road up a hill, and the road is 1,000 feet long. In that 1,000 feet it rises 100 feet. 100 divided by 1,000 is 0. 10, or 10%. It doesn't really matter if it is feet, meters, miles or kilometers -- if you know how far the road rises or falls in a given distance, you can calculate the percent grade by dividing the rise or fall by the distance.



If you are familiar with the trigonometry functions, you recognize that what you are calculating when you divide the rise by the distance is the tangent of an angle. Therefore, to calculate the number of degrees when you know the percent grade, you simply take the arctangent of the grade. So, if it is a 10% grade, you take the arctan of 0. 10 and you find that the angle is 5. 7 degrees.



Roads are probably measured in percent grade rather than degrees for two reasons:



You don't have to have a special calculator or trig tables to calculate percent grade

If you know the percent grade, it is very easy to calculate the distance you have risen or fallen simply by looking at your odometer. If you have coasted down a mile long hill on a 10% grade, you know you have fallen a tenth of a mile in vertical height.





Where did you find your "version" of grade?
 
I live on a private road with a 25% grade and a 70% turn halfway up the hill. I also have a 95 Toyota Landcruiser that I can lock the front and rear and center diff's. I use it when the weather gets real bad. My Dodge CTD will make it up the hill in the snow because of the weigh on the front in 4WD. I just don't want to drive it down the hill when it is snow covered. I don't think it would make the turn :D
 
My personal experience - your choice can be totally different.



In 88 bought the new body style 2WD GMC PU. Wanted a 4x4 but wimped out on the $$. Stuck that truck a hundred times.



In 93 traded for a Chevy 4x4 open diff. Stuck that 4x4 3 times.



In 97 traded for a Chevy 4x4 LSD. Never stuck that truck. With Goodyear 285 ATS tires, it would climb trees. Pulled a Ford with frozen rear brakes uphill in 6" of snow once.



In 1999 traded for 01 RAM Cummins 4x4 LSD 4. 10. Never was stuck and pulled 5th wheel trailer through some soft stuff. Pulled a Ford 2WD out of a 5 foot hole in the snow. Thought only a wrecker would be able to extract him. Son encouraged me to at least give it a try. One of my son's proudest moments! with stock Michelin M/S tires.



In November traded for 02 RAM cummins 4x4 LSD 4. 10. Pulled same 5th wheel through soft, wet, icy stuff no problem. Michelin stock A/S tires.



To make a long story short - this old boy ain't never going to have a truck that is not 4x4 LSD 4. 10.
 
I have been really inpressed with the Factory LSD in my 96 5sp. Between that and siped tires it takes a lots of snow to need 4x4. I have pulled lots of trailers through 8" on the level snow for hundereds of miles (middle of the night) and once rolling shifted out of 4x4 and never broke the rears loose.



I have 204K on mine and it is still as tight as new.



Tire wear ..... this is brand specific. I have had terrible wear on the factory Goodyear ATs, and so so wear on Dunlop ATs. But have got great even wear with Toyos. This is pulling hard.



Having had lots of HD Chevy trucks with their LSD, the Dodge one is much better.



jjw

ND
 
I wonder how many Rammers that have responded negatively towards the OE LSD, changed their diff. lube and arbitrarily added two bottles of Mopar friction modifier without any clue to what friction characteristics they may end up with???? :rolleyes: I've seen it written on this BB dozens of times.



BOTTOM LINE: the addition of friction modifier must be controlled... just to the point shudder is eliminated in slow, tight turns using a very small amount of throttle. Anything more and you are diminishing the effectiveness of your LSD... if you add more than that expect your LSD to act like an open differential.



Scott (Collins), the Tru-Trac is not recommended for high torque applications. Heck, when I made a call to them years ago they told me they won't warrant a Tru-Trac used with tires larger than a maximum of 32" (although I did have a friend with a built 350 in a early 80s GMC Sierra with 38" Monster Mudders wearing Tru-Tracs at both ends..... never had a problem!). However, I do know plenty of owners that had them grenade when used in the rear diff.
 
I think the LSD on my '96 works just great. I would not have a truck with a open diff. anymore. Got stuck too many times. I just pulled my friend's F350 (open diff. ) that was hooked up to a 6500 lb. chipper off some wet grass up a slight hill. One of his rear wheels was on the grass spinning and the other wheel was on concrete doing nothing. My truck pulled him out because I had traction on one wheel.
 
aj.com's info conflicting...

Originally posted by Thomas

Here is what I found at aj.com:



The percent grade is a simple calculation. Let's say you have a road up a hill, and the road is 1,000 feet long. In that 1,000 feet it rises 100 feet. 100 divided by 1,000 is 0. 10, or 10%. It doesn't really matter if it is feet, meters, miles or kilometers -- if you know how far the road rises or falls in a given distance, you can calculate the percent grade by dividing the rise or fall by the distance.



If you are familiar with the trigonometry functions, you recognize that what you are calculating when you divide the rise by the distance is the tangent of an angle. Therefore, to calculate the number of degrees when you know the percent grade, you simply take the arctangent of the grade. So, if it is a 10% grade, you take the arctan of 0. 10 and you find that the angle is 5. 7 degrees.






These two paragraphs conflict with each other. The first describes "rise over distance travelled", while the second describes "tangent of the angle".



"Rise over distance travelled" would be the sine of the angle, while "tangent of the angle" would be the rise over horizontal run.



sine = opposite / hypotenuse

tangent = opposite / adjacent



Assuming we overlay the "rise, run, distance" terms onto a right triangle:

opposite = rise

adjacent = horizontal run

hypotenuse = distance travelled



Therefore:

Rise / distance travelled = opposite / hypotenuse = sine

Tangent = opposite / adjacent = rise / horizontal run



Using aj.com's information quoted above, we get two different numbers for a 100% grade. The rise / distance travelled calculation says that this is a 90 degree angle. The tangent calculation says it's a 45 degree angle.



-Don
 
Trust me, I am always a surveyor. The 11. 8 degrees is correct. The 90 degrees/100 percent is incorrect. A 45 degree angle would be the 100 percent slope. The confusion on the previous post comes from the fact that a surveyor treats the world as if it was a plane surface (no topography). This is always true, even with a boundary survey. So, with the road example the linear distance traveled or the distance your truck traveled is not a plane distance because it would include topography. The 'run' or plane distance would be the horizontal distance between the 2 points. :)
 
Snowman-



I agree. I was just pointing out the inconsitencies in the website that kswieringa pointed us all to.



No confusion here. "Rise over run". dy / dx (i. e. "slope").



-Don
 
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