figuring grade steepness

[DPelletier]

The short answer: express the tangent of the angle as a percent.



The tangent of 4 degrees is a bit more than . 0699 or 6. 99%. A road that forms a 4 degree angle with the horizontal has a grade of about 7%. Since percent means "per hundred", if one travels on a road with a 7% grade, their elevation will increase or decrease 7 feet "per hundred" feet of horizontal movement. Their actual distance traveled on the road during this 7 foot change in elevation would be slightly more than 100 feet.



A road (trail?) that forms a 45 degree angle with the horizontal has a grade of 100%.

Excellent explanation. And of course to find out the actual distance travelled, we just need to use Pythagorean Theorum and use the A squared plus B squared = C squared; where A is the horizontal distance and B is the vertical distance.

Assuming you wanted to know the actual angle in the example above, you need to use; Tangent of the angle (theta) = Y over X where X is the horizontal distance and Y is the vertical distance.

See, all those years of Algebra and Trigonometry can be used on a regular basis! (well at least every couple of months or so )



Back to the original issue (I like the Chev country explanation!), While there are the odd side streets, etc. that are 30% grade and the Sonora Pass is 26% (although it's a stretch to consider that road a real highway), There are no interstates or major highways with a 30% grade. 6-8% would be considered a steep grade; 10-12% might be tolerated for very short sections. There's no way that the fellas friend went up a 30% grade, losing 5 mph, towing anything. If we really got industrious, we could probably figure out the hp required to perform as he has suggested if given the GVW of the package in question.



Cheers,

Dave

 

[Jumbo Jet]

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