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# How to convert treadmill percents to degrees?

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.

TIA

### Re: How to convert treadmill percents to degrees?

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, but hell we are after all
dealing with people in the exercise equipment, personal
trainers realm, so maybe it makes sense to THEM. LOL!

>>Evidently there is the tangent of an angle involved, and I
>>wonder why this is so.

DOH! "Evidently" ????

As the "wheelbase" and "rise" differ between machines then
the angle will also differ. Ooops, sorry, I forgot this IS
rocket science, unless of course we need to look at
Pythagoras' rather radical approach to things plane in a
new light.

I never cease to be amazed at the smarts (?????) of the
average American. Perhaps skipping dumb-ol Geometry that day
was not such a good plan in hindsight.

--

Kind regards,
Jenny and her tribe of survivors.

### Re: How to convert treadmill percents to degrees?

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
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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