The New York Times business section today ran an article about a new Oshkosh truck developed under a Pentagon contract. It stated that the truck could climb a 60-degree slope. I knew that was in error, and figured that somebody had meant to say "60 percent inclines."
Indeed that is the case. I left the newspaper in the coffee shop where I read it, but I see that the online version has been corrected, as you can see here: http://www.nytimes.com/2009/11/04/bu...20truck&st=cse.
What is percent slope? I've long held the belief that it is the ratio of vertical rise to distance traveled, which would be the sine of the angle (multiplied by 100 of course). But Wikipedia and my dictionaries define percent slope as the ratio of vertical rise to horizontal distance, which is the tangent of the angle. Anybody have a civil engineering textbook? I'll believe whatever that has to say on the subject.
This brings up a question as to what is the steepest slope a vehicle can manage. I've seen bulldozed firebreaks which almost defy credibility. How steep a slope can a bulldozer climb?
[edit] changed "slope" to "inclines," the word used in the NYT online article



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i wouldn't doubt it could climb a 100% slope (45 degrees) but you would need balls of steel to be driving it.
Crawlers are often asked to negotiate slopes deeper than they can dig. Sometimes the operators will push up a small tapered pad at the end so the dozer will dig steeper.