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Thanks.

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- Thread starter Brian P
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- #1

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Thanks.

- #2

Danger

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There's an obvious typo in your post, regarding the trailer weight. I certainly

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No, it is not a typo. I work in the coal mining industry, we haul giant size loads.

- #4

rcgldr

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Just stick with the .7 coeffiecient of friction. With a 42000 lb tractor, the maximum braking force is .7 x 42000 lbs = 29400 lbs. Total weight of the tractor and trailer is 42000 lbs + 110000 lbs = 152000 lbs. On a 8 degree slope, the component of gravity in the direction of the slope is sin(8 degrees) x 152000 lbs = 21154 lbs, which is less than the maximum braking force of 29400 lbs, so there's enough braking force to slow down and stop the tractor+trailer.

If the slope is > 11.1525 degrees, then the component of gravity in the direction of the slope is greater than the braking force and there tractor+trailer would accelerate even with maximum braking. You only have about 3 degrees of safety margin here so the slopes need to be made accurately.

This also assumes that the brakes don't fade. The brakes are performing 21154 lb ft of work for every foot braked, at 5 mph, this is 21154 lb x 5 mph x (5280 ft/mile) x (hour / 3600 sec) x (sec/550 lb ft) = 282 hp worth of heat generation.

If the slope is > 11.1525 degrees, then the component of gravity in the direction of the slope is greater than the braking force and there tractor+trailer would accelerate even with maximum braking. You only have about 3 degrees of safety margin here so the slopes need to be made accurately.

This also assumes that the brakes don't fade. The brakes are performing 21154 lb ft of work for every foot braked, at 5 mph, this is 21154 lb x 5 mph x (5280 ft/mile) x (hour / 3600 sec) x (sec/550 lb ft) = 282 hp worth of heat generation.

This information doesnt help unless you know the effective diameter (or radius) of the tires. At the front, there is 31500 lbs time the .7 coefficient of friction = 22050 lbs of force at the surface of the tires, or 11025 lbs per front tire. Braking torque is 622000/12 = 51833 ft lbs. This limits the tire radius to 51833 (ft lb)/11025(lb) = 4.7 ft for maximum braking, or 9.4 ft diameter. If only the braking force to control the tractor and trailer is considered, then the maximum allowable tire radius increases to (29040/21154) x 4.7 ft = 6.45 ft, or a diamter of 12.9 ft.I have 622,000 in-lb of braking in each axle.

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- #5

Danger

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Brian, I meant no offense about there being a typo, but there *is* one. What I didn't know was whether you put the comma in the wrong place, or if you left off a trailing zero. :tongue:

Since I thought that this was something to be done on a public road, I assumed the former. Sorry for the misunderstanding.

Since I thought that this was something to be done on a public road, I assumed the former. Sorry for the misunderstanding.

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- #6

rcgldr

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If the front tire diameter is less than 9.4 ft, then yes. The rears will lock up unless the braking bias is adjusted to balance with the load differential on the front and rear axles (31500lbs versus 10500lbs).If I slam on the brakes, will the wheels lock up?

Picture of a coal hauling truck, or a truck for those that think that a Hummer just isnt big enough to impress their neighbors:I work in the coal mining industry, we haul giant size loads.

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- #8

Danger

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I did not take offense Danger, sorry if I came across that way.

You didn't actually, but being a Canuck I always try to cover all of my don't-offend bases just in case. (Unless someone richly deserves it, in which case I'll offend him within an inch of his life. )

I'm not qualified to even begin helping you anyhow (never finished high-school). My post was more of just a greeting and to clarify something that I thought might confuse other readers.

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- #10

rcgldr

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The net (braking) force on the tractor + trailer is: 29400lbs - 21154 lbs = 8246lbs.

Mass of the system is 152000 lb mass.

Acceleration = - 8246lbs / 152000lbs = - .05425 g x (32.174 ft /(sec^2 x g) => - 1.7454 ft / sec^2

Initial speed is 5mph = 7.3333 ft / sec

Time to stop is (7.3333 ft / sec) / (1.7454 ft / sec ^2) = 4.2015 seconds.

Distance = V0 t + 1/2 A t^2 = (7.3333 ft / sec) (4.2014 sec) + 1/2 (-1.7454 ft / sec^2) (4.2015 sec)^2 = 15.4 feet

Mass of the system is 152000 lb mass.

Acceleration = - 8246lbs / 152000lbs = - .05425 g x (32.174 ft /(sec^2 x g) => - 1.7454 ft / sec^2

Initial speed is 5mph = 7.3333 ft / sec

Time to stop is (7.3333 ft / sec) / (1.7454 ft / sec ^2) = 4.2015 seconds.

Distance = V0 t + 1/2 A t^2 = (7.3333 ft / sec) (4.2014 sec) + 1/2 (-1.7454 ft / sec^2) (4.2015 sec)^2 = 15.4 feet

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