I think you meant 4 km\h @ 6% incline for 3 minutes.

(4000 \ 60) x 3 x 0.06 = 12m

No I meant while running a the pace of 1km every 4 minutes and 0 seconds (which I thought was written as 4:00/km when speaking pace) but your formula ignited the math functions in my brain. I can figure it out using the same logic.

Thanks so much!

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"Train so you have no regrets @ the finish line"

This is not exact but a good estimate would be-

Metric:
1000m / min/km * time(min) * grade
Ex) 1000m / 4min/km * 3min * .06 = 45m

or

English:

5280ft / min/mile * time(min) * grade
Ex) 5280ft / 6min/mi * 3min * .06 = 158ft

This is only an estimate, it works because the grades make relatively small angles. The smaller the grade, the more accurate the result. A grade of 15% only has an error of about 1%

What?

There is no error because the percent of grade is larger. 15% grade means a rise of 15' per 100'. A 6% grade means 6' rise in 100'. Mathematically there is no error.

sine(elevation % * 90 / 360) * distance ran, which in the case of 4% run for 5000 meters -- sin(4% * 90 / 360) * 5000 = 313 meters vertical and 4,687 horizontal.

First thank god for the mnemonic sohcahtoa - Sine = opposite / hypotenuse, Cosine = adjacent / hypotenuse, and Tangent = opposite / adjacent
Secondly thank god for excel so I can test whether I remembered my trigonometry accurately.

All is probably bunk as your treadmill is not that accurate in either distance or elevation.

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Swim - Bike - Run the rest is just clothing changes.

That's all fine if you are talking about degree of slope, but transportation generally references a change in elevation as percentage or grade which is rise\run. No trigonometric is required for the OP's question because he never stated he was running on a 6 degree slope, but rather a 6% slope..

Let me clarify ... I'm using the incline function on a standard treadmill set to 6. I assume that's a 6% incline but please correct me if I am wrong. I'm just trying to keep my elevation gains relatively accurate regardless if I am running inside or out.

Thanks to all who have helped! If there is something wrong that I've stated please do correct me.

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"Train so you have no regrets @ the finish line"

What Burnt Toast said.....100m forward, you are climbing 6 meters upwards. The percent grades are not angles and you need zero trig (no need to use sin(theta) math).

Burnt Toast is correct. if you want to go about seeing if the math works use this http://www.calcunation.com/...rcent-conversion.php

using that tool a 6% grade gives you an angle of 3.43 deg. Take the Tan of 3.43 and if you multiple by 100' to get the rise you get 6'. or 316' in a mile. of course the accuracy of your TM is not accounted for.

I ran trail marathon which had a mile with a 1,110 foot descent(47 ft up, but we'll ignore that). By Pythagoras sqrt(5280 * 5208 - 1110 * 1110) = 5,162 horizontal and 1,110 vertical make the mile that I ran.

By arcsin it is 12 degrees but by rise / run it is 21. Which number should I be quoting when talking about the pain my knees felt as I descended?

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Swim - Bike - Run the rest is just clothing changes.

No, trigonometry is required because % grade = rise/run or vertical distance/horizontal distance. You need trigonometry to figure out the run or horizontal distance. For low gradients the distance ran on the treadmill will be approximately the same as the horizontal distance, but as the gradient increases they diverge.

But on the dread-mill, the machine measures your horizontal distance. If you run 1000 meters at 10% grade, then you've run up 100 m vertical elevation. Don't see how any trig is needed:)

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"Anyone can be who they want to be IF they have the HUNGER and the DRIVE."