Coverting ground distance and elevation to grade percentage

I don't have a GPS so I'm trying to do this manually. If I have the ground distance, and the elevation change (as close as possible) from GMAP, how do I use the information to calculate the grade percentage? I've seen and read this post:
onlineconversion.com/forum/forum_1086616424.htm
but, I don't have the angle of the hill.

I've also seen and read this post:
gmap-pedometer.com/phpBB2/viewtopic.php?t=208
but, the final numers do not seem right.

The GMAP route, or hill, is this one:
gmap-pedometer.com/?r=645811
Using these two posts as reference, the final grade percentage comes out to 6%, which I think is wrong. Obviously, I need to manipulate the numbers I am getting in a different method than I am, or the GMAP values are wrong (which I doubt). So, for an elevation change of 31m, and a distance of 559m, what is the grade? On this hill, on my bike, I can start at the top, dead stop, and coast to 65kph by the time I hit the bottom. That's got to be more than 6% grade.

Any help is appreciated, thanks.

[Robert Fogt]

A rise of 31 meters 559 meters would equal:

31/559 * 100 = 5.55% grade

So that is correct, its about a 6% grade.

In a sense, your mind is not wrapping around it because its half what you would think. A 100% grade would be a 45° angle, but your mind is probably thinking 100% is straight up. So a 6% grade is twice as steep as your mind is thinking.

Now either I helped you understand it or confused you even more.

It would be great if this site had a calculator for converting grade expressed in terms of percentage, to grade expressed in terms of degrees.

I've found this formula for converting between the two, on the net but it took me a LONG time to find it. I thought it would be one of the calculators here (I always come running here for conversions). At least it got me to register.

degrees = arctan((percent grade)/100)
percent grade = tan(degrees) * 100

1% incline is a mere .57 degrees, 5% incline is 2.9 degrees, 10%
incline is 5.7 degrees, and 15% incline is 8.5 degrees.

Also, allow me to expand on Robert's pertinent point about percentage grade not having a ceiling of 100%.

Think of it this way, the percent grade is the rise as a percentage of the level distance.

Lets say you are about to travel up a hill from sea level. The summit of the hill is 50m above sea level. Lets also assume you are traveling in an easterly direction (you'll see why in a moment).

Assume you have GPS that is perfectly accurate (not invented yet) and at the summit it tells you that you've moved 50m east. You've moved the same amount vertically as you did horizontally. The horizontal distance was 100% of the vertical distance (i.e. 100% grade). If you draw 2 lines on a piece of paper that are the same length, one horizontal & one vertical and then draw a third line to join them up (forming a triangle) the angle from the horizontal to the line you drew last is 45º.

The sharpest hill you can have expressed in terms of degrees (angle) is 90º. Not really a hill actually, more like a vertical cliff face.

However if grade is expressed in terms of percentage it can be a lot more than 100%.

Lastly here's an image I dug up that shows a 10% grade.

EDIT: I see that because I'm a new user the system won't let me post pictures yet.

Does this help at all? I found it when I was searching for a tool to help me calucate the incline of a gradual hill for a marathon I'm running

csgnetwork.com/inclinedeclinegradecalc.html

I'm trying to calculate the distance between two locations. On my first point I have an elevation of 490.34 ft and on my second I have an elevation of 479.70ft. What do I have to do to calculate the distance?

[Dirtman]

Originally Posted by omar

I'm trying to calculate the distance between two locations. On my first point I have an elevation of 490.34 ft and on my second I have an elevation of 479.70ft. What do I have to do to calculate the distance?

The elevation difference is 10.64' but without more information, such as the angle between the two stations, you can't calculate the horizontal or slope distance. As your elevations are in 100ths of a foot, I'm assuming you shot it with a transit or theodolite, so set up on the first station and shoot the angle or . . . egads, simply measure it.

[JohnS]

Welcome Dirtman. Do you post at www.convert-me.com under the same name? You know your stuff. You'll find less spam and more interesting problems here.

[Dirtman]

Thanks John

Yes, the same Dirtman. Less spam here? LOL I haven't seen any spam at all yet. Compliments to the moderators. I spend more time deleting spam there than I do answering questions.

George

If I start at an elevation of 480.67 and drop 15 degrees vertically for 50', what elevation would I be at? Whomever answers, could you provide how you got this answer?