# training indoors for a hilly marathon - help!

July 29, 2011 8:57 PM Subscribe

I just ran 4.0 miles at a 4% incline on a treadmill. How much elevation gain does that equal, in feet?

I tried to figure it out by googling it, but I haven't found an actual equation where I can just plug in numbers, and the explanations seemed really complicated, involving arctan and stuff. Is someone out there math-savvy enough to figure it out, and give me an equation where I can just plug in my numbers for future runs?

I'm pretty sure my pace isn't relevant here, so I will threadsit for a while in case it's needed, but I'd rather not share because it's embarrassingly slow.

I tried to figure it out by googling it, but I haven't found an actual equation where I can just plug in numbers, and the explanations seemed really complicated, involving arctan and stuff. Is someone out there math-savvy enough to figure it out, and give me an equation where I can just plug in my numbers for future runs?

I'm pretty sure my pace isn't relevant here, so I will threadsit for a while in case it's needed, but I'd rather not share because it's embarrassingly slow.

Response by poster: Is it that simple, seriously?? Why all the crazy arctan and sin^-1 equations online?

Can someone else confirm this? Not that I don't believe you, but that elevation gain seems too good to be true!

posted by never.was.and.never.will.be. at 9:04 PM on July 29, 2011

Can someone else confirm this? Not that I don't believe you, but that elevation gain seems too good to be true!

posted by never.was.and.never.will.be. at 9:04 PM on July 29, 2011

Oh, also, if you ask google "4 miles * 4% in feet", it will do all the unit conversions and stuff for you

posted by aubilenon at 9:04 PM on July 29, 2011

posted by aubilenon at 9:04 PM on July 29, 2011

arctan is if you're looking for the angle. You don't care about the angle, you just want to know the total climb.

posted by aubilenon at 9:05 PM on July 29, 2011

posted by aubilenon at 9:05 PM on July 29, 2011

What crazy arctan and sin^-1 equations? Do you have a link?

posted by mr_roboto at 9:06 PM on July 29, 2011

posted by mr_roboto at 9:06 PM on July 29, 2011

Response by poster: Awesome, thanks! Don't judge my lack of math skills, I'm going to blame it on being super exhausted from climbing + running tonight.

posted by never.was.and.never.will.be. at 9:07 PM on July 29, 2011

posted by never.was.and.never.will.be. at 9:07 PM on July 29, 2011

Just because I like posting everything in tiny partial answers, here's some more: Wikipedia can confirm the term grade means what TheBones and I said. I promise I didn't edit it just to look good!

posted by aubilenon at 9:08 PM on July 29, 2011

posted by aubilenon at 9:08 PM on July 29, 2011

Response by poster: http://www.livestrong.com/article/38827-calculate-elevation-treadmills/

I didn't think I needed to know the angle, but everything I saw looked so complicated, and the only website that I saw that gave me the simple "4 miles x 4%" was yahoo answers, so I decided to come here to double check. Thanks!

posted by never.was.and.never.will.be. at 9:11 PM on July 29, 2011

I didn't think I needed to know the angle, but everything I saw looked so complicated, and the only website that I saw that gave me the simple "4 miles x 4%" was yahoo answers, so I decided to come here to double check. Thanks!

posted by never.was.and.never.will.be. at 9:11 PM on July 29, 2011

*Is it that simple, seriously?? Why all the crazy arctan and sin^-1 equations online?*

Look carefully; what you're putting into those equations is not a "4% grade" but a number of degrees. So the trigonometry has already been done for you. If you were told, say, that the hill was inclined at 2.3 degrees, then the grade would be arctan(2.3 degrees) = 0.0401, or 4.01% , and then you could do what aubilenon said in his first comment.

The reason that some of the formulas you're seeing involve arcsin, instead of arctan, is that some people get confused and define the grade to be Δh/l instead of Δh/d. (I'm using the labels from the Wikipedia article.) Some people are wrong, but it's easier to measure distances along the ground than imaginary horizontal distances, and for any reasonable angle d and l are really close together.

posted by madcaptenor at 9:24 PM on July 29, 2011

If you wanted an exact answer it would be:

height = distance * sin(arctan(grade))

The (.04 * distance) answer is only approximate because the distance you run is the hypotenuse, not the horizontal leg so the actual height is slightly less. It turns out that for grades of less than 15%, the approximation will be accurate to better than 1%.

posted by JackFlash at 9:52 PM on July 29, 2011 [2 favorites]

height = distance * sin(arctan(grade))

The (.04 * distance) answer is only approximate because the distance you run is the hypotenuse, not the horizontal leg so the actual height is slightly less. It turns out that for grades of less than 15%, the approximation will be accurate to better than 1%.

posted by JackFlash at 9:52 PM on July 29, 2011 [2 favorites]

Ahhhh! I think I see what's going on here. Your treadmill is measuring and reporting 4 miles as 4 miles in the direction of the treadmill (the hypoteneuse), not four miles in the x direction. So you do in fact want to take the sin(arctan(4%)) * 4.0 mi. So it's 844.12 feet instead of 844.8 feet. If you had been going up a steeper slope it would make a bigger difference.

arctan of a slope is its angle, and sin of an angle is its vertical component, is why I'm doing that there.

posted by aubilenon at 12:14 AM on July 30, 2011

arctan of a slope is its angle, and sin of an angle is its vertical component, is why I'm doing that there.

posted by aubilenon at 12:14 AM on July 30, 2011

Sorry to keep adding afterthoughts!

The other thing this means is that if you looked at those 4 hypothetical miles at 4% grade on a hypothetical map, you'd see that you came 16 ft of 4 miles away according to the map's scale - because those 4 miles are the diagonal distance, not the x component. cos(arctan(4%)) is the x component

posted by aubilenon at 12:26 AM on July 30, 2011

The other thing this means is that if you looked at those 4 hypothetical miles at 4% grade on a hypothetical map, you'd see that you came 16 ft of 4 miles away according to the map's scale - because those 4 miles are the diagonal distance, not the x component. cos(arctan(4%)) is the x component

posted by aubilenon at 12:26 AM on July 30, 2011

If you want it in concrete terms: you climbed the GE Building in NYC. Next time say hi to Jack Donaghy.

posted by rongorongo at 12:52 AM on July 30, 2011 [1 favorite]

posted by rongorongo at 12:52 AM on July 30, 2011 [1 favorite]

Elevation at end - elevation at start = 0 feet. Sorry. Treadmills are like that.

posted by trevyn at 2:17 AM on July 30, 2011

posted by trevyn at 2:17 AM on July 30, 2011

No, they're not like that. Running up a sloped treadmill is physiologically the same thing as running up a slope with a running-speed tailwind.

It's perfectly true that your gravitational potential energy at the end of a run up a real hill is greater than it was at the beginning (by your mass multiplied by gravitational acceleration multiplied by the height gain), and that this is not the case at the end of a treadmill run (no overall height gain). On the other hand (foot?) when you're running on a treadmill belt your legs are transferring kinetic energy to the belt, which doesn't happen when you're running on the road. In both cases, the work you do is the same.

The only time treadmill and road lose their equivalence is during those brief periods where the belt is changing speed, because then and only then does the belt's acceleration modify the gravitational field your body experiences.

posted by flabdablet at 3:13 AM on July 30, 2011

It's perfectly true that your gravitational potential energy at the end of a run up a real hill is greater than it was at the beginning (by your mass multiplied by gravitational acceleration multiplied by the height gain), and that this is not the case at the end of a treadmill run (no overall height gain). On the other hand (foot?) when you're running on a treadmill belt your legs are transferring kinetic energy to the belt, which doesn't happen when you're running on the road. In both cases, the work you do is the same.

The only time treadmill and road lose their equivalence is during those brief periods where the belt is changing speed, because then and only then does the belt's acceleration modify the gravitational field your body experiences.

posted by flabdablet at 3:13 AM on July 30, 2011

*If you want it in concrete terms: you climbed the GE Building in NYC. Next time say hi to Jack Donaghy.*

Or you ran from ground zero (street level) to the top of the GE building.

posted by pollex at 5:48 AM on July 30, 2011

This thread is closed to new comments.

(Incidentally, that means that there can be a > 100% grade; 100% is merely a 45 degree angle)

posted by aubilenon at 9:01 PM on July 29, 2011 [2 favorites]