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Could someone explain % in a climb

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Could someone explain % in a climb

Old 07-18-07, 09:11 AM
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Could someone explain % in a climb

I'm very new to cycling, starting this year at 44 and riding for fitness mainly but hoping to do some centuries and other club type rides. I've always enjoyed the TDF for what I think it is, a VERY demanding test of endurance, strength and mental capacity.

I know the % is a relation to length of climb and altitude gain. I was wondering at what degree of angle they are climbing for that distance? Is an 8% climb a climb of 8 degrees? Does this question even make sense to anyone?...lol. I told you I was new.
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Old 07-18-07, 09:12 AM
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vertical is 90deg or 100%.
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Old 07-18-07, 09:13 AM
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Rise/Run = % of climb.

i.e. 50 feet vertical over 1000 feet is a 5% grade.
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Old 07-18-07, 09:14 AM
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Originally Posted by dutret
vertical is 90deg or 100%.
incorrect

as the angle approaches vertical then % grade approaches infinity.

100% is 45 degrees. rise/run
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Old 07-18-07, 09:20 AM
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Originally Posted by baxtefer
incorrect

as the angle approaches vertical then % grade approaches infinity.

100% is 45 degrees. rise/run
I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.
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Old 07-18-07, 09:21 AM
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I understood it to be every 1ft elevation gain in 100ft distance travelled = your percentage. In other words 1ft elevation gain in 100ft distance travelled = 1% incline. 2ft elevation gain in 100ft distance travelled = 2% incline, etc.

Is this incorrect?
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Old 07-18-07, 09:27 AM
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Originally Posted by merlinextraligh
Rise/Run = % of climb.

i.e. 50 feet vertical over 1000 feet is a 5% grade.
That's the correct answer.
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Old 07-18-07, 09:30 AM
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Check this out . . . all you could want to know:

https://www.geocities.com/sidestreetluge/grade.html
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Old 07-18-07, 09:31 AM
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It depends if you use the sine function (opposite over hypotenuse. elevation gain over actual distance traveled) or if you use the Tangent function (opposite over adjacent. elevation gain over horizontal distance.)

In real world cases they come out very close to each other but in extreme cases they begin to differ a lot.
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Old 07-18-07, 09:34 AM
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Wikipedia has a pretty good explanation. I think it is more complicated. The answer previously given that a 45 degree angle is a 100% grade is correct.

https://en.wikipedia.org/wiki/Slope

See the section on calculating the slope of a road/railroad:

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Old 07-18-07, 09:35 AM
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Originally Posted by dutret
I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.
Wrong.
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Old 07-18-07, 09:36 AM
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This is an interesteing tid-bit from your site (https://www.geocities.com/sidestreetluge/grade.html) I did not know...


Q) Why use percent of incline vs. degree of angle?

A) The reason for using percent of incline instead of angle, is that percent gives a direct way to assess the effort required to move forward against the grade, whereas the angle in degrees does not readily reveal this information. For example; a 5% grade requires a forward force equal to 5% of the weight of the
object
(above and beyond the force it takes to overcome surface resistance on flat ground at the same
speed).
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Old 07-18-07, 09:39 AM
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Originally Posted by Keith99
Wrong.
According to the link previously given, they say
Measuring the distance along the surface of the road (c) instead of horizontally (b) gives practically
the same result for most road gradients up to 10%. For instance, a 20% grade (11.3 degrees), measuring
along the road surface gives a 19.6% grade (you must have a tangent handbook or calculator to properly
calculate these figures).
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Old 07-18-07, 09:41 AM
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I didn't know that, either (about % gradient being tied to the force necessary to climb the slope). It gives me a new perspective on the importance of resistance training.

Strength is good!
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Old 07-18-07, 09:42 AM
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Originally Posted by Turboem1
It depends if you use the sine function (opposite over hypotenuse. elevation gain over actual distance traveled) or if you use the Tangent function (opposite over adjacent. elevation gain over horizontal distance.)

In real world cases they come out very close to each other but in extreme cases they begin to differ a lot.

I started to add you can get a hell of a lot more complicated about it, but simply rise/run works pretty well for cycling purposes
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Old 07-18-07, 09:43 AM
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Originally Posted by fprintf
According to the link previously given, they say

no that just means they're close for normal grades.
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Old 07-18-07, 09:46 AM
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Originally Posted by Hammertoe
This is an interesteing tid-bit from your site (https://www.geocities.com/sidestreetluge/grade.html) I did not know...


Q) Why use percent of incline vs. degree of angle?

A) The reason for using percent of incline instead of angle, is that percent gives a direct way to assess the effort required to move forward against the grade, whereas the angle in degrees does not readily reveal this information. For example; a 5% grade requires a forward force equal to 5% of the weight of the
object
(above and beyond the force it takes to overcome surface resistance on flat ground at the same
speed).
I'm pretty sure that's not true.
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Old 07-18-07, 09:49 AM
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Wow, I had absolutely no idea it was this complicated. Makes sense if you think about it though.
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Old 07-18-07, 09:52 AM
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Originally Posted by dutret
I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.
It is usually calculated with the run being the length of the road, simply to make the math a little easier. For the angles on any sort of "normal" road though -- it simply doesn't make any signifigant difference to the final answer.
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Old 07-18-07, 09:55 AM
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It's simple geometry really. I usually use the pythagorean theorem to calculate the slope, IE on a 1 mile , 500 foot climb

a^2+b^2=c^2
a^2+250000=27878400
a=5256

rise/run = b/a = 500/5256 = ~0.095 or 9.5% gradient
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Old 07-18-07, 09:56 AM
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Originally Posted by KevinF
It is usually calculated with the run being the length of the road, simply to make the math a little easier. For the angles on any sort of "normal" road though -- it simply doesn't make any signifigant difference to the final answer.

I was under the impression that percent grade was synonymous with gradient not just conveniently close around 0 which means that it would be calculated that way and the rise/run method would be the convienent approximation.
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Old 07-18-07, 09:56 AM
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25% means get off and walk!
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Old 07-18-07, 09:57 AM
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Originally Posted by toucci
It's simple geometry really. I usually use the pythagorean theorem to calculate the slope, IE on a 1 mile , 500 foot climb

a^2+b^2=c^2
a^2+250000=27878400
a=5256

rise/run = b/a = 500/5256 = ~0.095 or 9.5% gradient
gradient is most definitely not calculated this way.
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Old 07-18-07, 09:58 AM
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Originally Posted by dutret
gradient is most definitely not calculated this way.
Ok could you correct me then?
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Old 07-18-07, 10:01 AM
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no I'm wrong.
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