How to convert treadmill percents to degrees?

(Halterb) said:

Quoted message said:

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!

Quoted message said:

Quoted message said:

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.

Quoted message said:

On 25 Jun 2004 10:29:11 GMT, [email hidden]

(Halterb) said:

Quoted message said:

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.