Silly question about roads and percent grade.

I see these signs here in the US, like next 2 miles, 13% grade. Anyone know what that means in terms of degrees? I would assume a horizontal road flat on open ground would be zero degrees but what represents 100% grade? 45 degrees? 90 degrees? Something else?

Originally posted by sonhouse
I see these signs here in the US, like next 2 miles, 13% grade. Anyone know what that means in terms of degrees? I would assume a horizontal road flat on open ground would be zero degrees but what represents 100% grade? 45 degrees? 90 degrees? Something else?

I think 13% means that after 100 meter you have changed your elevation with 13 meter. If the slope is only 10 meters, then you get 1.3 meter upwards.
0% is flat, 100% would be vertical.

Originally posted by sonhouse
I see these signs here in the US, like next 2 miles, 13% grade. Anyone know what that means in terms of degrees? I would assume a horizontal road flat on open ground would be zero degrees but what represents 100% grade? 45 degrees? 90 degrees? Something else?

The mile length is the hypotenuse of a (right) triangle (let's call that side 'c'😉, in your example,
2 miles. The grade percentage is the ratio of the height of that triangle to the length of it (let's
call those sides 'b' and 'a' respectively).

To figure out the angle, you know that the tangent of the angle is the ratio of those two sides
(opposite over adjacent), so you can take the arctangent of that ratio (in your example, .13),
and get 7.407 degrees.

Thus a 100% grade would be 45 degrees (a/b=1.00, because the right triangle is also isosceles).

Nemesio

So we have two different opinions here! I like version #2 simply because the road I ride on sometimes, called Mountain road here, says specifically, the grade is 13% for 2 3/4 miles and it is a rather gentle downhill and just intuitively seems like it couldn't be more than 7 degrees. If FF is right, the slope would be 14 degrees, which seems a bit steep.
I wonder if there are any roads around that are actually 45 degrees, like I wonder what the slope is in those San Fransisco numbers? Ever drive on those, man they are STEEP!

Originally posted by sonhouse
So we have two different opinions here! I like version #2 simply because the road I ride on sometimes, called Mountain road here, says specifically, the grade is 13% for 2 3/4 miles and it is a rather gentle downhill and just intuitively seems like it couldn't be more than 7 degrees. If FF is right, the slope would be 14 degrees, which seems a bit steep.
I ...[text shortened]... nder what the slope is in those San Fransisco numbers? Ever drive on those, man they are STEEP!

I might be wrong of the 100% example.
I see it as a ratio, but in the extrem, well, then we have another problem, the power of the engine... 🙂

Originally posted by sonhouse
So we have two different opinions here! I like version #2 simply because the road I ride on sometimes, called Mountain road here, says specifically, the grade is 13% for 2 3/4 miles and it is a rather gentle downhill and just intuitively seems like it couldn't be more than 7 degrees. If FF is right, the slope would be 14 degrees, which seems a bit steep.
I ...[text shortened]... nder what the slope is in those San Fransisco numbers? Ever drive on those, man they are STEEP!

Yeah, that one windy street is bizaare. It's the steepest street in the world or something.

Originally posted by AThousandYoung
Yeah, that one windy street is bizaare. It's the steepest street in the world or something.

Lot of car chase movie scenes filmed on that road, cars making sparks on the road as they fly up and hit the ground then magically keep on going getting away from the bad guys...

Originally posted by FabianFnas
I think 13% means that after 100 meter you have changed your elevation with 13 meter. If the slope is only 10 meters, then you get 1.3 meter upwards.
0% is flat, 100% would be vertical.

Almost. It's not the hypotenuse you compare the elevation to, but the other leg of the triangle.

So, a 13% grade over a 100 meter hypotenuse rises 12.89 meters. The other leg is 99.17 meters.

Nemesio

Originally posted by Nemesio
Almost. It's not the hypotenuse you compare the elevation to, but the other leg of the triangle.

So, a 13% grade over a 100 meter hypotenuse rises 12.89 meters. The other leg is 99.17 meters.

Nemesio

Well I assume that would be proportional, it doesn't have to be a 100 meter hypot, it could be 50 meters and the other two #'s just divided in two.

Originally posted by sonhouse
Well I assume that would be proportional, it doesn't have to be a 100 meter hypot, it could be 50 meters and the other two #'s just divided in two.

Precisely. I was just working out the answer to the problem posed by FabianFans. Given the
ratio that the two legs have to each other, in a right triangle, the angle must be 7.407 degrees,
no matter what the hypotenuse is. But, if you want to find the actual change in elevation, you
need to know either the horizontal distance traveled or the hypotenuse.

Nemesio

Generally, when I see those grade % signs, it's accompanied by a maximum speed limit for
trucks weighing over a certain amount. I suspect it's because the increased angle can lead to
losing traction with the ground and slipping.

How much more traction does a body need to have to keep from slipping when it's on a hill
10% steep? 20%?

Nemesio

Originally posted by Nemesio
Generally, when I see those grade % signs, it's accompanied by a maximum speed limit for
trucks weighing over a certain amount. I suspect it's because the increased angle can lead to
losing traction with the ground and slipping.

How much more traction does a body need to have to keep from slipping when it's on a hill
10% steep? 20%?

Nemesio

If I screw this one up, my high school physics teacher is going to kill me from beyond the grave.

For a truck moving at constant velocity applying its brakes on a flat surface, we have the following:

Fn = Fg
mu = Fb / Fn

where:

Fn = normal force
Fg = gravitational force
Fb = force due to braking
mu = coefficient of friction (either rolling or static, whichever is appropriate)

Therefore, the maximum braking force that can be applied without slipping is Fb = mu*Fn.

For a truck initially moving at constant velocity applying its brakes on a downward slope with angle of elevation theta, where theta = Tan^(-1)(grade in % ), we have:

Fn = Fg*cos(theta)
Fa = Fg*sin(theta)

where Fa = force of acceleration due to the angle of elevation. Also:

Fb = mu*Fn = mu*Fg*cos(theta)

This is the maximum amount of braking force that can be applied, which is strictly less than the amount that can be applied on a flat surface. Also, the braking becomes more ineffective and the push from the ramp becomes more effective as the angle of elevation increases, according to the formula for the resultant force (in the "down-the-ramp" direction):

Fr = Fa - Fb = Fg*sin(theta) - mu*Fg*cos(theta) = Fg*(sin(theta) - mu*cos(theta))

Letting Fr = 0 (the break-even point where the truck starts going down the hill regardless of braking power), we can see that:

sin(theta) = mu*cos(theta)
mu = tan(theta)

So for a given mu, an angle of elevation exceeding Tan(-1)(mu) will result in horrible, firey death. 🙂

Just tried working this out on a spreadsheet, and got a nice simplification (which I should have seen in the first place, and is probably the reason why civil engineers use grade % in the first place).

Since the maximum angle of elevation before disaster, theta(max), is Tan(-1)(mu), and the maximum grade % is just tan(theta(max)), we end up with:

maximum grade % = mu

Just checking some figures on the web. Apparently for dry roads, mu is about 0.7, but on wet roads it can be as low as 0.4. Of course, there are other issues with hard-braking vs. soft-braking, stopping distance, etc... but it seems that as long as the grade never goes about about 20% we should be fine.

Originally posted by Nemesio
Generally, when I see those grade % signs, it's accompanied by a maximum speed limit for
trucks weighing over a certain amount. I suspect it's because the increased angle can lead to
losing traction with the ground and slipping.

How much more traction does a body need to have to keep from slipping when it's on a hill
10% steep? 20%?

Nemesio

The problem with trucks isn't so much about slipping as it is over-heating the brakes. On long grades they'll need to run slower and in a lower gear, so that their "braking" comes from engine compression rather than riding the brakes all the way down the grade.

Originally posted by leisurelysloth
The problem with trucks isn't so much about slipping as it is over-heating the brakes. On long grades they'll need to run slower and in a lower gear, so that their "braking" comes from engine compression rather than riding the brakes all the way down the grade.

Ah, then you haven't heard about the new braking system invented recently: Explosive bolts that drive anchors into the road with ropes that play out and slow down the trucks in an emergency, all they need are two such devices, one on either side and the truck slows down evenly. On curves there is a computer that follows the curve and plays out the lines differentially so it handles those as well.