# How to figure out vertical climb on a treadmill?

April 14, 2016 6:50 AM Subscribe

Because I am clueless at geometry, at least I think it's geometry, could someone please show me how to figure out how many vertical feet I would have climbed if I hiked for a mile on a treadmill set at 15%?
Thank you.

oh, 15%!! that's 8.5 degrees. and the sine of that is 0.15. so 264 yards per mile. sorry!

(obs: for small angles like this, sin() is approx eq to tan() which is why sine is 0.15 which matches the tangent (15% is a tangent of 0.15)).

posted by andrewcooke at 7:01 AM on April 14, 2016

(obs: for small angles like this, sin() is approx eq to tan() which is why sine is 0.15 which matches the tangent (15% is a tangent of 0.15)).

posted by andrewcooke at 7:01 AM on April 14, 2016

784 feet.

If you were to move 100 feet horizontally on a 15% grade, you would rise 15 feet.

A mile is 5280 feet. 15% of that is 792 feet. However, your mile isn't horizontal, it is on the 15% grade. So imagine a big skinny right triangle. The mile is the hypoteneuse, so the actual horizontal distance walked is less than a mile. About 5,228 feet. 15% of that is 784 feet.

posted by yesster at 7:02 AM on April 14, 2016 [4 favorites]

If you were to move 100 feet horizontally on a 15% grade, you would rise 15 feet.

A mile is 5280 feet. 15% of that is 792 feet. However, your mile isn't horizontal, it is on the 15% grade. So imagine a big skinny right triangle. The mile is the hypoteneuse, so the actual horizontal distance walked is less than a mile. About 5,228 feet. 15% of that is 784 feet.

posted by yesster at 7:02 AM on April 14, 2016 [4 favorites]

The full formula (including conversion from grade to degrees) is:

Height Climbed = sin(tan^-1(slope percent/100))*(distance travelled)

You can plug this into Wolfram Alpha to get your height, just replace the numbers as appropriate. The one I linked would be for walking 100 units at a 15% grade.

posted by one of these days at 7:04 AM on April 14, 2016 [1 favorite]

Height Climbed = sin(tan^-1(slope percent/100))*(distance travelled)

You can plug this into Wolfram Alpha to get your height, just replace the numbers as appropriate. The one I linked would be for walking 100 units at a 15% grade.

posted by one of these days at 7:04 AM on April 14, 2016 [1 favorite]

Based on the link below 792 feet:

http://www.livestrong.com/article/525801-how-to-calculate-an-elevation-gain-for-a-treadmill/

posted by tman99 at 7:09 AM on April 14, 2016

http://www.livestrong.com/article/525801-how-to-calculate-an-elevation-gain-for-a-treadmill/

posted by tman99 at 7:09 AM on April 14, 2016

As I showed, there is no need for using geometric formula at all.

Just use the 15%.

The livestrong article isn't correcting for the fact that the treadmill's measured mile is the hypoteneuse of the right triangle.

792 would be correct if the horizontal distance were one mile. But it is not. The horizontal distance is less than one mile.

Imagine the grade were 100%. Your treadmill says you walked one mile. But that mile is the angle of the right triangle, not the bottom. So the vertical and horizontal distances are both 3,733 feet. (using pythagorean theorem)

posted by yesster at 7:16 AM on April 14, 2016 [2 favorites]

Just use the 15%.

The livestrong article isn't correcting for the fact that the treadmill's measured mile is the hypoteneuse of the right triangle.

792 would be correct if the horizontal distance were one mile. But it is not. The horizontal distance is less than one mile.

Imagine the grade were 100%. Your treadmill says you walked one mile. But that mile is the angle of the right triangle, not the bottom. So the vertical and horizontal distances are both 3,733 feet. (using pythagorean theorem)

posted by yesster at 7:16 AM on April 14, 2016 [2 favorites]

No trig necessary.

A grade of 15% means that you went x feet horizontally and .15x vertically.

By the Pythagorean theorem, you went √ [x^2 + (.15x)^2] = 1.0112x feet diagonally.

That's 5280 feet, so 1.0112x = 5280, so x = 5221.6 feet.

So your vertical climb, which we saw is .15x, is 783.2 feet.

posted by dfan at 7:18 AM on April 14, 2016 [1 favorite]

A grade of 15% means that you went x feet horizontally and .15x vertically.

By the Pythagorean theorem, you went √ [x^2 + (.15x)^2] = 1.0112x feet diagonally.

That's 5280 feet, so 1.0112x = 5280, so x = 5221.6 feet.

So your vertical climb, which we saw is .15x, is 783.2 feet.

posted by dfan at 7:18 AM on April 14, 2016 [1 favorite]

But you didn't climb a single vertical foot, so all estimates are WRONG (at least if you are trying to estimate how far or fast you could hike in the real world).

The reason going uphill is hard is because you are doing work lifting your body against the pull of gravity (w= mass*gravity*height)

On a treadmill, it's complicated: depending on how you walk, you may be basically keeping your center of mass stationary and moving your legs a bit extra. So, the actual amount of extra work you did by using an inclined treadmill is probably somewhere between zero and "same as if you were walking up a mountain". I'm sure someone has calculated this before.

posted by soylent00FF00 at 5:43 PM on April 14, 2016 [2 favorites]

The reason going uphill is hard is because you are doing work lifting your body against the pull of gravity (w= mass*gravity*height)

On a treadmill, it's complicated: depending on how you walk, you may be basically keeping your center of mass stationary and moving your legs a bit extra. So, the actual amount of extra work you did by using an inclined treadmill is probably somewhere between zero and "same as if you were walking up a mountain". I'm sure someone has calculated this before.

posted by soylent00FF00 at 5:43 PM on April 14, 2016 [2 favorites]

Thanks, everyone!

posted by holdenjordahl at 8:46 AM on April 15, 2016

posted by holdenjordahl at 8:46 AM on April 15, 2016

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(that sounds a lot. maybe i have made a mistake?)

posted by andrewcooke at 6:59 AM on April 14, 2016