Kinda math/climbing/training type question...
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Kinda math/climbing/training type question...
For some reason the math is escaping me at the moment but we just got a treadmill at home and I wanted to figure out elevation gain based on the incline I set it too... It goes up to 15% and I can set my own program so I want to set it for a 1000 foot gain and then go from there. Can I enlist the math types out there? If I do a mile at a 15%, for example, how many feet elevation gain is that?
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Re: Kinda math/climbing/training type question...
Fletch wrote:880 ft. Sound right?
Wouldn't it be 5280 feet * 0.15 = 792 feet ?
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Re: Kinda math/climbing/training type question...
Fletch wrote:880 ft. Sound right?
Percent grade: just take 15% of the run. 0.15*5280 ft = 792 ft
For 1000 ft gain at 15%, 0.15*X = 1000 ft, X = 6,667 ft = 1.3 mi.
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Re: Kinda math/climbing/training type question...
jomagam wrote:Fletch wrote:880 ft. Sound right?
Wouldn't it be 5280 feet * 0.15 = 792 feet ?
So 1000 feet of elevation would be 1000/0.15/5280 = 1.26 miles
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Re: Kinda math/climbing/training type question...
Technically I think the mile would be the hypotenuse in this triangle because that's where the treadmill track is.
If I'm doing my math and understanding treadmills correctly you have fifteen feet up for every one hundred feet horizontal. So taking the arctangent of 15/100 you have an angle of incline of about 8.53 degrees. Take the sine of that angle and multiply it by your hypotenuse, 5280 feet, and you have a vertical gain of 783 feet.
I could be wrong, I don't know much about how they measure inclines on treadmills.
If I'm doing my math and understanding treadmills correctly you have fifteen feet up for every one hundred feet horizontal. So taking the arctangent of 15/100 you have an angle of incline of about 8.53 degrees. Take the sine of that angle and multiply it by your hypotenuse, 5280 feet, and you have a vertical gain of 783 feet.
I could be wrong, I don't know much about how they measure inclines on treadmills.
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Re: Kinda math/climbing/training type question...
the answer is just a tad over 783 feet. woops! Robzilla beat me to it.
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Re: Kinda math/climbing/training type question...
climbing_rob wrote:the answer is just a tad over 783 feet. woops! Robzilla beat me to it.
There you go, 100% of Robs agree. Can't be wrong.
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Re: Kinda math/climbing/training type question...
I'm assuming the 15% means that you are going up .15 units for every 1 unit you go horizontally. Then we can construct a right triangle with base x, height .15x, and hypoteneuse 5280. Using the Pythagorean theorem, I calculated .15x to be about 783 ft.
Alternatively, if you want to know how far you need to run in order to do 1k ft vertical, we need to find the hypoteneuse of a right triangle having height 1000 and base 1000/0.15. Using the Pythagorean theorem, I got 6741' or about 1.3 mi.
woops, Robzilla beat me too!
Alternatively, if you want to know how far you need to run in order to do 1k ft vertical, we need to find the hypoteneuse of a right triangle having height 1000 and base 1000/0.15. Using the Pythagorean theorem, I got 6741' or about 1.3 mi.
woops, Robzilla beat me too!
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Re: Kinda math/climbing/training type question...
Robzilla wrote:Technically I think the mile would be the hypotenuse in this triangle because that's where the treadmill track is.
If I'm doing my math and understanding treadmills correctly you have fifteen feet up for every one hundred feet horizontal. So taking the arctangent of 15/100 you have an angle of incline of about 8.53 degrees. Take the sine of that angle and multiply it by your hypotenuse, 5280 feet, and you have a vertical gain of 783 feet.
I could be wrong, I don't know much about how they measure inclines on treadmills.
I have no experience in how treadmills work, and suspect they're nowhere near accurate enough for this to matter, but defining the gradient in percentages to mean the ratio of the two legs doesn't sounds right to me. IMO a 100% grade should be mean vertical, not 45 degree as you calculate it. By your logic anything steeper than 45 is over 100% grade; that just sounds silly to me.
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Re: Kinda math/climbing/training type question...
Nope, opinion or not, "100% grade" means that for every foot of horizontal, you go a foot of vertical. That's the definition of percent grade. Vertical has infinite "percent grade" (1 foot of vertical for 0 feet horizontal--> 1/0 = infinity).jomagam wrote:IMO a 100% grade should be mean vertical, not 45 degree as you calculate it. By your logic anything steeper than 45 is over 100% grade; that just sounds silly to me.
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Re: Kinda math/climbing/training type question...
Apparently 100% of Robs here are also aerospace engineers. Kooky. Rocket science came in handy again!
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Re: Kinda math/climbing/training type question...
climbing_rob wrote: Nope, opinion or not, "100% grade" means that for every foot of horizontal, you go a foot of vertical. That's the definition of percent grade. Vertical has infinite "percent grade" (1 foot of vertical for 0 feet horizontal--> 1/0 = infinity).
I stand corrected.
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