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Re: 10% on a treadmill for 3 miles... vertical is? [pedaller]
Ok, my take...

Edited: When I was writing this post the diagram was not screwed up, but when posted my leading spaces were dropped.
.
. l /\
. l the rise, or the vertical gain, we'll call this "X"
. l \/
.____________________x

<horizontal distance > we know this is ten times the rise, so we can call it 10X. FYI, the grade in this example is easy to deal with, but if it was something other than a 10% grade we'd just divide 1 by the grade to come up with our coefficient for X. For example, if it was a 6% grade, 1/.06 is 16.66, so the sides would be X and 16.66X.

actual path that the runner traveled was along the dotted line, it is 3 miles.

now to solve for X:

Pythagorean theorem states: a**2 + b**2 = C**2

c must be the side of the triangle that is opposite from the right angle of the triangle. This is the same thing as saying C is the hypotenuse. In my crappy diagram above, where the "x" is located is where the right angle (90 degrees) is.

A is one side, doesn't matter which, I'll say it's X.

B must be the other side, 10x.

so x**2 +(10x)**2 = 3**2

solving the algebra...

x**2 + 100X**2 = 9 (note the 10 has been squared and shows as 100, the exponent now only applies to x)
101X**2 = 9 (since in both cases x is raised to the second power, algebra allows us to add the coefficients, when X just sits there with no number in front of it the coefficient is 1).
x**2 = 9/101 (I divided each side by 101)
x = square root of 9/101 (I've taken the square root of each side).
x = approximately .2985
x is stated in miles.

I don't think I'm full of crap, but I'm not 100% sure.

OK, and here is a real connection to all of this bs and triathlons. We can see that .2985 is very close to .3 ( the three tenths of a mile the original poster estimated the vertical gain to be). The reason it's just barely short of .3 is because the horizontal was not 3 miles, but rather the hypotenuse was 3 miles. If the horizontal would have been 3 miles than the vertical gain would have been exactly .3 miles. If these were the distances the hypotenuse would have to be approximately 3.015 miles. In other words if the runner ran 3.015 miles at a 10% grade, the vertical gain would be .3 miles.

Observation: When the grade is low (say 10% or less) the horizontal gain is very close to the actual running distance.

How does this connect to triathlons?

Consider the chaos at the beginning of the swim. At the start everyone is headed to the first buoy. People swimming on top of people. Let's say it's 1/2 mile to the buoy, 880 yards. Now let's say you moved 10% of that distance, approximately 90 yards to the side and swam from that point to the buoy. Not nearly the choas. And, as demonstrated above, you've added very little to your swim distance. In the example above, moving 10% of the distance to the side increased the travel distance from 3 to 3.015. In the half mile example, if you moved 90 yards to the side, your distance to the buoy would be approximately 884 yards instead of 880 yards.

Therefore... It's worth it to move way to the side to avoid the battle field of swimming chaos.

Steve

Last edited by: StevePupel: Jan 6, 11 20:59

Edit Log:

  • Post edited by StevePupel (Big Pines) on Jan 6, 11 20:59