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Grade Calculations(11 posts)

Grade CalculationsEl Guapo
Jun 5, 2003 10:08 AM
Someone please help me understand. I know it is rise over distance. But, is the distance based on the travelled distance (hypoteneuse) or the linear distance. For reference: a 337 ft rise over the course of 2640 ft. travelled = 12.8% grade - OR - a 337 ft. rise over linear distance of 2303 ft. = 14.6% Seems an insignificant discrepency, but just curious none the less.
re: Grade CalculationsDougSloan
Jun 5, 2003 10:38 AM
Technically, it's rise of the horizontal distance, but nearly everyone calculates it from linear distance, as that's not as easy. I guess the steeper the hill, the more the error, right?

Is there an equation for conversion of linear to true percentage?

Doug
re: Grade CalculationsEl Guapo
Jun 5, 2003 10:56 AM
Here's how I calculated it: Given that this is a triangle with at least one known angle of exactly 90 degrees, Pythagoreans theory should work ( a2 + b2 = c2 ). Rise = 337 ft. Distance travelled (hypoteneuse) = 2640 ft. 337/2640 = 12.8% To find the horizontal (linear) distance I used the following: a = horizontal, b = rise, and c = Hypoteneuse. {a + 337 = 2640} {a = 2640 - 337} {a = 2303} Then, 337 (Rise)/ 2303 (horizontal distance) = 14.6%. I think this is correct.
re: Grade CalculationsWW
Jun 5, 2003 11:38 AM
I think your math is a little off. The best thing to do is check your work. If a=337, b=2303, and c=2640, then rerun the equation. However you will find that:
a2 + b2 = 5417378 = c2.
The square root of c2 (5417378) = 2327.5, not 2640.

The correct equation is:
a2 + b2 = c2, therefore,
b2 = c2 - a2
(b2)1/2 = (c2-a2)1/2, where 1/2 is used to designate
square root
Therefore,
b = (6969600 - 113569)1/2
b = 2618
Go back and try a=337, b=2618, and c=2640.

So the slope is 337/2618 = 12.87%.
re: Grade CalculationsEl Guapo
Jun 5, 2003 11:57 AM
I stand corrected. 12.87% still seems steep to me.
re: Grade CalculationsMellowVelo
Jun 5, 2003 10:47 AM
The CORRECT side of calculation is linear distance, but on most grades the difference is so small so usually it's no big deal, but the steeper the grade gets, the bigger the discrepancy. In your case, a 1.8% difference seems pretty big to me.

Spence
clarificationMellowVelo
Jun 5, 2003 10:55 AM
I should clarify.. the actual distance travelled is not the correct side (hypoteneuse) to base the calculation on. Linear distance, being the horizontal plane of the climb w/no rise is the correct side to measure. If you are using your bike computer to figure the distance, that will be the hypoteneuse of course. It sounds like you know the math though.
re: Grade CalculationsEl Guapo
Jun 5, 2003 11:03 AM
That 1.8% makes a huge difference. I don't feel so bad struggling up that slope with only a 39/23 as my friendliest gear.
Here's another fun one for you.El Guapo
Jun 5, 2003 11:22 AM
203 ft. rise over 1056 ft. travelled distance. = 19.22%
If I calculate using the linear: a + 203 = 1056
a = 1056 - 203
a = 853
203/853 = 23.798% !!!!!
YIKES!
By the way, the rise numbers are based on geographical surveys conducted by one of my friends companies. They do topographic maps. The distance travelled is based from point-to-point survey positions, so should be as accurate as can be. I know these two climbs are short, but these are only two of no less than 8 available, hardly used, well-paved climbs of similar gradient here in Austin.
Update as per WWEl Guapo
Jun 5, 2003 12:08 PM
a=Linear distance, b=rise, c=travelled distance(hypoteneuse)
a=?, b=203, and c=1056
a(a) + 203(203) = 1056(1056)
a(a) + 40609 = 1115136
a(a) = 1115136 - 40609
a(a) = 1074527
a = 1036.59

203/1036.59 = 19.583% actual gradient

Thanks to WW for the assistance. I had to go back and rememeber how to calculate square roots without the assistance of a scientific calculator.
that's alot of numbers.JS Haiku Shop
Jun 5, 2003 12:21 PM
they all make my head hurt.

dude, i'm diggin' your handle.

-J