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. |
vertical is 90deg or 100%.
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Rise/Run = % of climb.
i.e. 50 feet vertical over 1000 feet is a 5% grade. |
Originally Posted by dutret
(Post 4877427)
vertical is 90deg or 100%.
as the angle approaches vertical then % grade approaches infinity. 100% is 45 degrees. rise/run |
Originally Posted by baxtefer
(Post 4877436)
incorrect
as the angle approaches vertical then % grade approaches infinity. 100% is 45 degrees. rise/run |
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? |
Originally Posted by merlinextraligh
(Post 4877432)
Rise/Run = % of climb.
i.e. 50 feet vertical over 1000 feet is a 5% grade. |
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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. |
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.
http://en.wikipedia.org/wiki/Slope See the section on calculating the slope of a road/railroad: http://upload.wikimedia.org/math/e/7...31073cb9cb.png http://upload.wikimedia.org/math/1/f...9c0a7bfc9a.png |
Originally Posted by dutret
(Post 4877470)
I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.
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This is an interesteing tid-bit from your site (http://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). |
Originally Posted by Keith99
(Post 4877595)
Wrong.
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). |
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! |
Originally Posted by Turboem1
(Post 4877560)
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 |
Originally Posted by fprintf
(Post 4877618)
According to the link previously given, they say
no that just means they're close for normal grades. |
Originally Posted by Hammertoe
(Post 4877599)
This is an interesteing tid-bit from your site (http://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). |
Wow, I had absolutely no idea it was this complicated. Makes sense if you think about it though.
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Originally Posted by dutret
(Post 4877470)
I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.
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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 |
Originally Posted by KevinF
(Post 4877735)
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. |
25% means get off and walk!
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Originally Posted by toucci
(Post 4877766)
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 |
Originally Posted by dutret
(Post 4877788)
gradient is most definitely not calculated this way.
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no I'm wrong.
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