Re: How to convert treadmill percents to degrees?

From: Dr Chaos (mbkennelSPAMBEGONE_at_NOSPAMyahoo.com)
Date: 06/28/04 

Date: Mon, 28 Jun 2004 03:12:35 +0000 (UTC)

Jenny3kids@msn.net <Jenny3kids@msn.net> wrote:
> On 25 Jun 2004 10:29:11 GMT, halterb@aol.com (Halterb) wrote:
>
>>>Could anyone give me some information on treadmill incline (grade) conversions
>>>from percents (as in Bruce protocol) to degrees? I've found some conflicting
>>>information--i.e 1 percent equals .59 degrees, or 1 percent equals 1.73
>>>degrees. Evidently there is the tangent of an angle involved, and I wonder why
>>>this is so.
>
> Errr, how long is a piece of string?
>
> The percent option will be independent of the "wheelbase" and "rise"
> of the incline mechanism, whereas the degrees will be an actual angle.
>
> You are asking if a short cow can jump as far and high as a tall dog.
>
> It is simple trigonometry to calculate the angle when given the
> "wheelbase" (i.e. support points) and the "rise" (distance from zero
> incline)
>
> Tan(Angle) = Rise / Wheelbase
>
> Percent will be anything you want it to be. I assume by percent they
> would mean that flat is 0% and the full extension (max incline) of the
> raising mechanism is 100%. Unless you are using trigonometry and real
> measurements, then the percent is meaningless,

It is probably meant in terms of "vertical rise is X percent of
horizontal". Think about the signs that say "7% downgrade ahead" on
the road. That is not in reference to any particular machine.

Hence reporting the slope as X% is reporting the tangent of the slope angle:

1% grade: tan(theta) = 0.01
which yields theta = 0.573 degrees.

That's my guess.

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