15% incline = how much rise in elevation in one mile?

"15%" = 13.5 deg since a "100%" grade would be 90 deg.

zdfgadf wrote:

"15%" = 13.5 deg since a "100%" grade would be 90 deg.

why are you bringing degrees into this?

zdfgadf wrote:

"15%" = 13.5 deg since a "100%" grade would be 90 deg.

No, 100% grade = 45 deg

15% grade = arctan(0.15) = 8.53 deg

You're fixing "15% incline" = rise/horizontal, and varying what "one mile" refers to.

I varied what "15% incline" referred to, and fixed what "one mile" referred to.

hyp. = 5280', rise/horizontal run = .15 -> 783' vert = 5280sin(arctan(.15))

hyp. = 5280', rise/road dist = .15 -> 792' vert = .15*5280

Your framing is better I guess.

It cracks me up how even the simplest of subjects on anything math/physics related inevitably becomes a long debate.

We're not quite up to airplanes on treadmills here but...

none. your feet remain exactly 0' above the road.

Not enough information to solve.

Are we taking elevation as sine of the path or as a distance coincident with the mass center of the earth?

What is your 5k PR?

Do you look relaxed?

Need more info wrote:

Not enough information to solve.

Are we taking elevation as sine of the path or as a distance coincident with the mass center of the earth?

What is your 5k PR?

Do you look relaxed?

Right now I look relaxed. But I am not running a 5k PR at this time. I am floating in a hot tub.

Also, my 5k PR is better than all hobby joggers and worse only than those on PEDs.

Finally, I flew over Wisconsin once but my plane was not on a treadmill at the time.

And I currently identify as someone who thinks this should be enough information.

Cracker Me Up wrote:

Need more info wrote:

Not enough information to solve.

Are we taking elevation as sine of the path or as a distance coincident with the mass center of the earth?

What is your 5k PR?

Do you look relaxed?

Right now I look relaxed. But I am not running a 5k PR at this time. I am floating in a hot tub.

Also, my 5k PR is better than all hobby joggers and worse only than those on PEDs.

Finally, I flew over Wisconsin once but my plane was not on a treadmill at the time.

And I currently identify as someone who thinks this should be enough information.

You've just added more variables to the system. There are not enough constraints. Even assuming a simplified Newtonian model of this "mile" won't bring the precision required to satisfy any self-respecting master of science.

15%.

Trig Correction wrote:

zdfgadf wrote:

"15%" = 13.5 deg since a "100%" grade would be 90 deg.

No, 100% grade = 45 deg

15% grade = arctan(0.15) = 8.53 deg

Most would consider a sheer wall (90% vertical incline) to be a 100% grade, no?

Hj457 wrote:

Trig Correction wrote:

No, 100% grade = 45 deg

15% grade = arctan(0.15) = 8.53 deg

Most would consider a sheer wall (90% vertical incline) to be a 100% grade, no?

No. Percent grade is 100*(rise/run). If a road is at a 45 degree angle, it rises one foot for every foot of run. 100*(1/1)=100%.

Faulty definition then. What retard decided that?

Hj457 wrote:

Faulty definition then. What retard decided that?

John Vanderselzhien.

Hj457 wrote:

Trig Correction wrote:

No, 100% grade = 45 deg

15% grade = arctan(0.15) = 8.53 deg

Most would consider a sheer wall (90% vertical incline) to be a 100% grade, no?

When the amount risen is equal to the amount traveled sounds like a good way to define the term "100% grade".

thank you I never knew how the grade of a road was defined.

Visiting LetsRun is a learning experience on most days and very often I learn stuff I did not even know I should know.

Wrong 45* is 100%

Nope, 45°