Re: How to convert treadmill percents to degrees?

Jenny3kids_at_msn.net
Date: 06/25/04 

Date: Fri, 25 Jun 2004 06:27:40 -0700

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.

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