Is an incline setting of 15% on a treadmill the same as a road at 15% grade?
It should be off by at least 1% since that is required to make the treadmill closer to running outside on flat land so 1% tread = 0% road, but how about 15%?
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The intent is that the incline of a treadmill matches a grade. However, as noted with your comment on 0% incline, there's no air resistance, and some people have noted that inclines aren't always accurate.
The Fitness.StackExchange link provided by @Adam is great and probably the most complete when answering your question directly.
The actual calculations for treadmill v road grade are as follows.
This link talks about calculating grade from a treadmill.
The treadmill grade is basically a measure of the height distance for every 100 (insert unit of measurement here) horizontal distance. e.g. A one in 100 gradient = 1%, and a rise of 15 meters for every 100 meters is a 15 % grade.
Here is an excellent answer about road grade from a biking forum.
What does percent grade mean on a road?
When you travel through the mountains, you often see signs that say things like "Trucks check brakes -- 10% grade" or "6% grade -- Trucks use right lane only." These numbers obviously have something to do with the steepness of the road, but their exact meaning is a mystery to most drivers. The grade of something is simply a measure of its rise over its run. To understand rise and run, it helps to think of the hill as a big right triangle (a triangle with a 90-degree angle), like this:
The rise is the length of side B, or the height of the hill. The run is the length of side A, the horizontal measure of the hill at ground level. So, if you rose 100 feet over a horizontal distance of 1,000 feet, rise over run would equal 100 divided by 1,000, or 0.1. To get the percent grade, you simply multiply by 100, which gives you 10%. It doesn't matter whether you use feet, meters, miles or kilometers -- if you know how far the road rises in a given horizontal distance, you can calculate the percent grade. To calculate percent grade exactly, you need to figure out the horizontal distance traveled (A). Since you know B and C, you can calculate A using the Pythagorean theorem -- "the square of the hypotenuse is equal to the sum of the squares of the other two sides," or:
C2 = A2+ B2
This means that:
A = SquareRoot (C2 - B2)
If you had driven 1,000 miles down the road and risen 100 miles, the horizontal distance would be the square root of 990,000, which is approximately 994.99 (rounded up). So what good does all of this do you? First of all, the percent grade gives you a relative sense of how steep the hill is. If you've climbed a hill designated as having a 5% grade, for example, you'll know approximately what to expect from any other 5% grade hill.
Notice that the same measurements occur both in treadmill and road form. The downside to both become the accuracy of the equipment and the real world application.
Treadmills can be off for any number of reasons. Because the grade is measured via distance and height -> if the odometer or incline motor are off, the calculations produce the same on-screen Grade% while the treadmills can be at very different angles.
In the real world, a 10% grade over a 5 mile long climb (in the mountains, for example) is not always a change of 528 feet per mile (5,280 feet). Sometimes there are longer somewhat flatter sections and then more steep sections. On my drive back from college there were several "severe grade" warnings where you wouldn't be through the full decent of the mountain but you'd be climbing a hill again because of the lay-of-the-land.