% Grade Question

Quoted post said:

Originally posted by tanggoman
How does one relate a % grade to an angle in degrees. If someone says its 15%-20% grade, how does one convert it in degrees? Thanks!

grade = elevation change / distance

angle = arcsin (grade)

Get ready to be underwhelmed with your findings.

Quoted post said:

Originally posted by rollers
grade = elevation change / distance

angle = arcsin (grade)

Get ready to be underwhelmed with your findings.

Sorry for the stupid question but how do u do an ArcSin in a calcuator? 😕

Quoted post said:

Originally posted by tanggoman
Sorry for the stupid question but how do u do an ArcSin in a calcuator? 😕

because that depends on the calculator you're using I can't give you an answer to that question

If by chance you could use a spreadsheet program such as Microsoft Excel, the cell formula you'd use would look like this:

= DEGREES(ASIN(grade))

where grade = elevation change / distance

Quoted post said:

Originally posted by tanggoman
How does one relate a % grade to an angle in degrees. If someone says its 15%-20% grade, how does one convert it in degrees? Thanks!

A 45 deg. slope is a 100% grade. Use 45 as your constant and multiply by percent grade expressed as a decimal.

ex. convert a 20% slope to degrees

45degs. times .20 = 9deg

ex. convert a 9deg slope to a percent

9deg/45 =.2=20%

Try a few, you'll see that it's fool proof.

Thanks!

Quoted post said:

Originally posted by rollers
grade = elevation change / distance

angle = arcsin (grade)

Get ready to be underwhelmed with your findings.

Is it not arctan?? 😕

Do they measure rise / distance travelled, rather than rise / horizontal distance??

Quoted post said:

Originally posted by pineapple
Is it not arctan?? 😕

Do they measure rise / distance travelled, rather than rise / horizontal distance??

You are RIGHT! I had it as distance travelled but the correct dimension to use is horizontal distance. Therefore arctan is the right operator.

correcting my Excel formula: =DEGREES(ATAN(grade))

Here is a good link to explain it fully:
http://draco.acs.uci.edu/rbfaq/FAQ/9.3.html

Shame on me for not researching this before opening my big keyboard.🙄

Quoted post said:

Originally posted by rollers
You are RIGHT! I had it as distance travelled but the correct dimension to use is horizontal distance. Therefore arctan is the right operator.

correcting my Excel formula: =DEGREES(ATAN(grade))

Here is a good link to explain it fully:
http://draco.acs.uci.edu/rbfaq/FAQ/9.3.html

Shame on me for not researching this before opening my big keyboard.🙄

lol I'm all confused now cos I thought for a minute that your original theory might be right...

🙁

I vaguely remember hearing somewhere that the roadsign people are stupid and measure it using sine... but maybe thats just over here in the uk (apparently we're not that clever after all)

Anyway tbh its pretty irrelevant 😄

Quoted post said:

Originally posted by pineapple
...
I vaguely remember hearing somewhere that the roadsign people are stupid and measure it using sine... but maybe thats just over here in the uk (apparently we're not that clever after all)

Anyway tbh its pretty irrelevant 😄

what's funny about this is that I'd had the same thought. I even considered leaving in a comment about civil engineers (the road builders here in the US) but I took it out. LOL

Ditto on the irrelevant. Grade is the right way to say it, certainly my preference. Have you ever done any of these calculations? It's amazing how a big grade is actually a quite small angle. 12% grade equates to less than 7deg. 7deg! That doesn't sound all that hard to climb now does it?

20% grade = 11.5deg

What do we need these small chainrings for?

Thanks rollers, it was my layman way to do the conversion. After studying trig on the net I see the err of my ways. Now... If I wanted to put satelite in orbit around mars...........

I think, for the angle calculation one should use:

theta = arcsin(elevation gained/dist travelled),
where "theta" is slope angle.

The (elevation gain)/(horizontal translation) will give you (%grade), and tan(theta).

theta = arctan(%grade)

So, if you wanna get wacky:
sin(theta) = sin[arctan(%grade)] = elevation gained/dist travelled

So, if you live in the SF Bay Area and you go on a typical ride here's how it might work out for you:

The ride: 3000ft elevation GAIN, over a 30mile ride.

AVERAGE %grade:

3000ft/(30 x 5280ft) = sin(theta)

theta = arcsin[3000ft/(30 x 5280ft)]

AVERAGE %grade = tan(theta) = tan[arcsin(1000/52800)] = 1.9%

Don't forget there's downhill, too. So, let's just call it 3.8% AVERAGE grade on the climbs of this typical 30-mile ride.

My rationale is that you'll never know what the horizontal distance traveled is. You ride on the hypoteneuse of the triangle so you can get that dimension from your cycling computer, and the elevation gain can be had from a topo or an altimeter (gps sucks for this, btw). It's more relevant to a rider to know how far they will have to ride to gain a certain amount of altitude. The horizontal distance translated can only be judged from a satellite or a map. I guess you could use map information to calculate a grade and that grade number puts it in perspective for a rider. I guess what I'm trying to get at is that you could easily calculate grade with an altimeter and a cycling computer using the arcsin calculation.

Good luck figuring out horizontal distance travelled from the seat of your bike. Please let me know if you can do that. I'd like to know how it could be done.

[edited for my rusty trig and added example calc.]

Quoted post said:

Originally posted by Xsmoker
Now... If I wanted to put satelite in orbit around mars...........

Hmm, let's see... I can sustain 232 watts for 1 hour, 201 for several hours..... Now, how much work would be required to get me and my bike free of Earth's gravitational pull and onto an intercept path with Mars?

Gotta go do some internet research. With just a few bits of information I bet it'd be possible to figure out by what year I could get my own personal payload to Mars. I better plan my "launch" date carefully so as to arrive in time for that year's TdM.

A Ciclo Master 434 computer is a humbling device when you look down while gasping up that "Wall" of a hill and see 12% grade!
;-)

Quoted post said:

Originally posted by Pat Lee
The horizontal distance translated can only be judged from a satellite or a map. I guess you could use map information to calculate a grade and that grade number puts it in perspective for a rider. I guess what I'm trying to get at is that you could easily calculate grade with an altimeter and a cycling computer using the arcsin calculation.

Good luck figuring out horizontal distance travelled from the seat of your bike. Please let me know if you can do that. I'd like to know how it could be done.

[edited for my rusty trig and added example calc.]

I'll handle this one rollers. Since you know the distance traveled and the gain in altitude you can use the pythagorean theorem to calculate the horizontal distance.. But this isn't really necessary with angles less than 10 degs.

distance traveled2 - elevation gain2 = horizontal distance2

The twos indicate the square.

Quoted post said:

Originally posted by Xsmoker

distance traveled2 - elevation gain2 = horizontal distance2

The twos indicate the square.

Uh yeah, Pythagoras. That b$stard. You're right it doesn't matter much whether one uses sine or tangent, my point is theoretical grade calculations should use sine.

There are definitely places here that you would have to use the sine calculation anyway. There are several short rises where the grades are rated above 15%. I know one is rumored to be 26%, but you won't catch me measuring the hypoteneuse with my cycle computer on that one.

Oh yeah, and to the Pythagorean theorem calculation of the horizontal distance translated. I can't do that from my bicycle seat to a good precision. I can't do square roots in my head.

Quoted post said:

Originally posted by rollers
what's funny about this is that I'd had the same thought. I even considered leaving in a comment about civil engineers (the road builders here in the US) but I took it out. LOL

Ditto on the irrelevant. Grade is the right way to say it, certainly my preference. Have you ever done any of these calculations? It's amazing how a big grade is actually a quite small angle. 12% grade equates to less than 7deg. 7deg! That doesn't sound all that hard to climb now does it?

20% grade = 11.5deg

What do we need these small chainrings for?

As a Consulting Surveyor I thought I could confirm that :

Angle of incline = ArcTan Grade (where grade = vertical distance / horizontal distance) - so a 20% grade = 11.31 Degrees.

For those of you who still don't understand, check out the attached table !

This help ?? 😄

Quoted post said:

Originally posted by cpa
As a Consulting Surveyor I thought I could confirm that :

Angle of incline = ArcTan Grade (where grade = vertical distance / horizontal distance) - so a 20% grade = 11.31 Degrees.

For those of you who still don't understand, check out the attached table !

This help ?? 😄

That is precisely what I think the OP was asking for. Not my amateur prattle for sure!

--Pat.

Quoted post said:

Originally posted by Pat Lee
That is precisely what I think the OP was asking for. Not my amateur prattle for sure!

--Pat.

Hehe.

And don't forget ....... If you're going up a hill, your c-computers are measuring the SLOPE distance (i.e. the Hypotenuse of the triangle).

so if you know the height difference (from a map or altimeter) you can also calculate the following :

Angle of Incline (or decline) = ArcSin (dH / Ds).

OR

Grade = dH / Square Root [Ds(squared) - dH(squared)]

where ;
dH = Difference in height
Ds = Slope distance

**This is, of course, assuming that it is a constant grade !!!

Fwoooaarrr...... feel like ridin' a big peak now !

Confused yet ???!!!😄 :p 😉 (sheesh, I've been soundin' like a squid [geek])

Yeah, see? that was my point, too. You're riding the hypoteneuse. The horizontal can really only be calculated. It's actually theoretical, in the sense that it can only be calculated from other measurements. Unless of course you have access to tunneling equipment and follow the projection of the road on a horizontal plane underneath. :o

I think this horse is dead now. Next?