Help with Hill %



Jaguar27

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Sep 19, 2003
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OK, I know I'm stoopid...but...

If I ride .217 miles and gain 95 feet, what would the % be?

Thanks for your help!!
 
Jaguar27 said:
OK, I know I'm stoopid...but...

If I ride .217 miles and gain 95 feet, what would the % be?

Thanks for your help!!
looks like about 8%

convert miles to feet then divide rise by run, convert to percentage
so
.217 miles = 1,145.76 feet (used and online converter)
95/1,145.76 =.083 = 8.3%
 
Thank Eden!
I did it the wrong way around, divided the run by the rise and got 12%...

So it's only 8% ??? wow!!

Thanks again!!

Eden said:
looks like about 8%

convert miles to feet then divide rise by run, convert to percentage
so
.217 miles = 1,145.76 feet (used and online converter)
95/1,145.76 =.083 = 8.3%
 
Here's a super easy way to do this if you are online. Just go to Google and enter:
95 feet/0.217 miles in percent
Google is smart enough to do the conversions and give you the same correct answer posted by Eden.
 
Chance3290 said:
Ah, it was my understanding that there would be no math involved in this forum.
Chase. Chevy. Gerald Ford.

Thanks for the memory and the laugh.
smile.gif
 
The trick is to climb an 8% grade at a fast enough pace that the atmospheric pressure drops quickly enough to make your ears pop.

This feat gets harder and harder every year...
 
Cheers for that, I luv it!!
And thanx for all the other replies...you Guys are fantastic..

The reason I was asking is that I use a, what I now know, is an 8.3% Climb for an interval...I also use it to gauge whether I'm getting better or worse, I turned 50 a few Days ago and luckily I *seem* to be getting better...

So here's me...on a good Day, 8.1 mph when I get to the top...more if a Mountain Lion is chasing me...but I can manage a 60 cadence, which is 25 lower than my *normal* Cadence...

The problem is though, I'm gonna Climb Haleakala in March, my new Mate, Donny Arnault (He doesn't know it yet) told me it's an average of 6%...that doesn't sound too bad untill I take into consideration that it's over 38 mls up to 10,023 ft...

I'll probaly have to fill my Water Bottles with Guiness to accomplish this feat!!

I have a 13/29 on the rear and a, I think, 38t on the front...so I reckon I'll beable to spin up...maybe...I'll climb like a Goat, albeit an old one..but there ya go...

So I reckon I'd better spend all my rides between now and then going up hills, which are easy to find around here...

Also, I have a century coming up in a week and a half, so I reckon if I drag a Roll of Chainlink Fence around the entire route I should be good to go...right??

Cheers then!!


Scarpelli said:
The trick is to climb an 8% grade at a fast enough pace that the atmospheric pressure drops quickly enough to make your ears pop.

This feat gets harder and harder every year...
 
Eden said:
looks like about 8%

convert miles to feet then divide rise by run, convert to percentage
so
.217 miles = 1,145.76 feet (used and online converter)
95/1,145.76 =.083 = 8.3%
Chance3290 said:
Ah, it was my understanding that there would be no math involved in this forum.
Now for the real math lesson. There was a similar thread a while back, where it was pointed out that there is a difference between the true run and the measured distance traveled. This is a good example of why, at the grades most people can climb, it does not make any difference.

The true grade is the rise divided by the run, as Eden correctly stated. The rise is 95 feet. The distance you traveled is the hypotenuse. The run would be the third leg of the triangle that connects the starting point to the point 95 feet below the finishing point. If you traveled 0.217 miles, that is 1,145.76 feet (again, as correctly stated by Eden), but the run is only 1141.81 feet (the square root of (1145.76 feet squared minus 95 feet squared)). Thus, the "true" grade is 8.32009% instead of 8.29144%.

Other than geometry teachers, would anyone care about a difference that small? Now, if you were able to climb a 45 degree incline (100% grade), the difference between using the actual distance traveled rather than the true run is big. Using the distance traveled, you would calculate the grade as 70.07107%, but it is really 100.00000%.

End of math lesson - back to cycling.
 
Thanks Rick, now I can say "Blimey, these 8.32009% climbs feel a bit tough these Days don't they" to my fellow Riders....

RickF said:
Now for the real math lesson. There was a similar thread a while back, where it was pointed out that there is a difference between the true run and the measured distance traveled. This is a good example of why, at the grades most people can climb, it does not make any difference.

The true grade is the rise divided by the run, as Eden correctly stated. The rise is 95 feet. The distance you traveled is the hypotenuse. The run would be the third leg of the triangle that connects the starting point to the point 95 feet below the finishing point. If you traveled 0.217 miles, that is 1,145.76 feet (again, as correctly stated by Eden), but the run is only 1141.81 feet (the square root of (1145.76 feet squared minus 95 feet squared)). Thus, the "true" grade is 8.32009% instead of 8.29144%.

Other than geometry teachers, would anyone care about a difference that small? Now, if you were able to climb a 45 degree incline (100% grade), the difference between using the actual distance traveled rather than the true run is big. Using the distance traveled, you would calculate the grade as 70.07107%, but it is really 100.00000%.

End of math lesson - back to cycling.