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Old 09-22-05, 08:15 PM   #1
TheRCF
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Hill calculation formula

Okay, I now have some data on my hills, but I need to be able to translate it into a grade percentage.

If I understand it correctly, going up hill for 5 miles with a 1 mile increase in elevation, would result in an angle of 20 degrees. If the increase were 2.5 miles, the angle would be 45 degrees. I know this isn't quite right since you should compare the horizontal distance, which would be a little less than the diagonal distance, but I suppose for bike angles of climb, that is not a major factor.

But how do you get the percentage of the grade? Isn't that different?
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Old 09-22-05, 09:02 PM   #2
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To calculate grade it

number of verticle (units) (usually feet)
---------------------------------------
number of horizontal (units) (same as above)


so for a 1 mile stretch of road if you gain 528 feet

528 ft
you have grade = ---------------- = 10% pretty steep climb
5280 ft


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Old 09-22-05, 09:03 PM   #3
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Actually your angle is off...

Take the inverse tangent of the elevation divided by the horizontal to get the angle. To have a 45 degree angle (or 100% gradient), your elevation gain will equal your horizontal distance. In your example, arctan(1/5) = 11.31 degrees.

Gradient is simply rise divided by run. So 1 mile / 5 miles = 0.2 = 20% gradient = steep.
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Old 09-22-05, 09:27 PM   #4
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You lost me at "inverse tangent". Is there a field expedient (idiots) formula to use?
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Old 09-22-05, 09:31 PM   #5
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When you ride 5 miles up a hill, 5 miles is not the horizontal distance!

The angle (not the grade) of the slope is ARCSIN(1 mile/5 miles) = 11.537 degrees

The grade is just the TAN of the angle. TAN(11.537) = .204 or 20.4%

I uploaded a spreadsheet to do this a while back if you can find it.
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Old 09-22-05, 09:35 PM   #6
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Okay, let me see if I understand this.

Gradient, which is a PERCENTAGE is the vertical distance divided by the horizontal distance?

But this formula gives the ANGLE: arctan(verticla/horizontal) = a result in DEGREES?

If I have that right, I have two questions. First, which figure, percentage or degrees is most often used in hill descriptions (I think it is percentage)?

Second, since I'm trying to set up formulas for calculating the hills, I ran into a problem. Using Excel spreedsheet, it seems that "arctan" is not a term it recognizes. I found Atan, AtanH, and Atan2, but none of them seems to give a result like you show for the 1 mile vertical on a 5 mile ride. Any ideas on what the formula should be in excel?
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Old 09-22-05, 09:41 PM   #7
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Quote:
Originally Posted by TheRCF
Okay, let me see if I understand this.

Gradient, which is a PERCENTAGE is the vertical distance divided by the horizontal distance?
Correct.

Quote:
But this formula gives the ANGLE: arctan(verticla/horizontal) = a result in DEGREES?
Correct

Quote:
If I have that right, I have two questions. First, which figure, percentage or degrees is most often used in hill descriptions (I think it is percentage)?
Gradient.

Quote:
Second, since I'm trying to set up formulas for calculating the hills, I ran into a problem. Using Excel spreedsheet, it seems that "arctan" is not a term it recognizes. I found Atan, AtanH, and Atan2, but none of them seems to give a result like you show for the 1 mile vertical on a 5 mile ride. Any ideas on what the formula should be in excel?
ATAN returns the answer in radians. To convert into degrees, multiply by 180 / PI().
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Old 09-22-05, 09:44 PM   #8
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Originally Posted by JavaMan
When you ride 5 miles up a hill, 5 miles is not the horizontal distance!
You are right, but it makes a difference in our case of 0.2 degrees and 0.4% gradient (and will always make a negligable difference in biking terms since the elevation gain <<< traveled distance unless you're doing sick, sick climbs). For doing quick and easy gradient calcs on the fly, rise over (fake) run is good enough for government work!
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Old 09-22-05, 09:48 PM   #9
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OK, I found the post I was talking about - I did not upload the spreadsheet after all, but just posted the formula in spreadsheet form.

This is the correct formula. RISE is in feet, DISTANCE is in miles.

%GRADE = TAN(ASIN(RISE/(5280*DISTANCE)))*100

Put it into a spreadsheet for future use.
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Old 09-22-05, 10:02 PM   #10
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Ok, JavaMan is getting hardcore with the significant digits. And I can dig that. So here is a way to use Excel to get down to the decimal points.

You need to know the (D)istance you travel up the hill (well will call that T and put that number in cell A1)
You also need to know the (E)levation gain from the bottom of the hill to the top of the hill (we will call that E and put that number in cell B1)


The formula for percent grade in cell C1 should look like this =B1/(SQRT((A1*A1)-(B1*B1))) format this to percent to 2 places

My previous example of 582ft elevation on a 1 mile hill should give the following results:

T=5280 ft in cell A1 // A1=5280
E=528 ft in cell B1 // B1=528

Using the formula above Cell C1 = 10.05%

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Old 09-22-05, 10:07 PM   #11
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Quote:
Originally Posted by JavaMan
OK, I found the post I was talking about - I did not upload the spreadsheet after all, but just posted the formula in spreadsheet form.

This is the correct formula. RISE is in feet, DISTANCE is in miles.

%GRADE = TAN(ASIN(RISE/(5280*DISTANCE)))*100

Put it into a spreadsheet for future use.

Shoot! Beat out by JavaMan.

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Old 09-22-05, 10:07 PM   #12
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Quote:
Originally Posted by JavaMan
OK, I found the post I was talking about - I did not upload the spreadsheet after all, but just posted the formula in spreadsheet form.

This is the correct formula. RISE is in feet, DISTANCE is in miles.

%GRADE = TAN(ASIN(RISE/(5280*DISTANCE)))*100

Put it into a spreadsheet for future use.
Okay, I have SirScott's formula working and we already know there is a small built-in error.

So the question becomes: Is your formula more accurate?

One of the nice things about computers is that once set, even a complicated formula can just be copied and pasted as necessary so I might as well go with whatever is best.

But while waiting for comments on the most accurate figures, I sure appreciate all the help.

Just saw another post by Fatboy but I'll have to check it out more later.
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Old 09-22-05, 10:25 PM   #13
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Quote:
Originally Posted by TheRCF
Okay, I have SirScott's formula working and we already know there is a small built-in error.

So the question becomes: Is your formula more accurate?

One of the nice things about computers is that once set, even a complicated formula can just be copied and pasted as necessary so I might as well go with whatever is best.

But while waiting for comments on the most accurate figures, I sure appreciate all the help.

Just saw another post by Fatboy but I'll have to check it out more later.
JavaMan will have to explain his formula. My formula is more accurate than the simple elevation/distance. You can see in my example the real answer is 10.05% (rounded) and the close answer was 10%. Believe me, .05% wont make any difference, but as Sir Scott said, the steeper the hill, the more inaccurate it becomes. But trust me, if you can climb a 20% grade, you are the man in my books.

Steve
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P.S. My formula calculates the real horizontal lentgh using c squared = a squared + b squared, where the distance up the hill is c and the elevation gain is a, and the real horizontal distance is b. Then I used the elevation / real horizontal distance to get the "real" grade.
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Old 09-22-05, 11:33 PM   #14
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Okay, using Fatboy's formula - hopefully I finally did it right after a number of errors in my setup kept giving weird results - the steepest hill I think I've climbed covers three short blocks and the steepest part is 8.63. The average for all three is 7.07.

The distance is just 1065 feet. I've climbed it a few times with my old bike, which has a triple crank, but I have yet to make it with my newer bike (double crank). I've gotten half way up, but that's about it. I'm so slow, I have to worry about falling over so I can't make lock into the pedals - I'm afraind I'd start to fall and not be able to unclip successfully.
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Old 09-23-05, 12:11 AM   #15
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Quote:
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...the steepest hill I think I've climbed covers three short blocks and the steepest part is 8.63. The average for all three is 7.07.
8 and 7 percent is pretty step. I did 2.5 miles on the back side of the Royal Gorge that I think averaged 10% with a few sections around 12. It was hard. One technique I saw other people use (this was on an MS-150 ride) was zig-zaging. I couldn't make that work. I just used my smallest gear. 30x25. I have a pretty standard road tripple.

Good luck with those hills.

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Last edited by Fat Boy Biker; 09-23-05 at 11:07 AM.
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Old 09-23-05, 07:57 AM   #16
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My biggest hill is 10.7% over about three blocks (using Google Earth to determine distance and elevation change) and I have to zigzag up it, even on my electric bike! I can't climb standing on the pedals and my heart rate went through the roof when I zigzagged up it on my Trek, even with a 32 granny gear.
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Old 09-23-05, 11:36 AM   #17
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To figure grade at one point. Use a 4' level and the # of inches from the street to the
bottom of the level (with the bubble leveled) divided by 48 = % slope.

If it is 4" from the street to the bottom of the level your equation is: 4/48 = .08333 which
would be 8.3% slope.

If you are using a 2' level you would divide by 24
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Old 09-23-05, 01:26 PM   #18
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My biggest hill is 10.7% over about three blocks (using Google Earth to determine distance and elevation change) and I have to zigzag up it, even on my electric bike! I can't climb standing on the pedals and my heart rate went through the roof when I zigzagged up it on my Trek, even with a 32 granny gear.
Before getting the Delorme Topo USA software, I tried that - but a bunch of places, the numbers made no sense. I would check a place right along the beach and it would say I was 20 or more feet high! Even over the ocean - you know, like SEA LEVEL - it would say I was well above that. Places that I knew should be going downhill would show increasing elevation!
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Old 09-23-05, 01:55 PM   #19
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Well, I just got one of these


so I'll doublecheck my Google Earth results.
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Old 09-23-05, 03:25 PM   #20
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Quote:
Originally Posted by CastIron
You lost me at "inverse tangent". Is there a field expedient (idiots) formula to use?
So all that would be a 'NO' then?
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Old 09-23-05, 04:22 PM   #21
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Quote:
Originally Posted by Fat Boy Biker
8 and 7 percent is pretty step. I did 2.5 miles on the back side of the Royal Gorge that I think averaged 10% with a few sections around 12. It was hard. One technique I saw other people use (this was on an MS-150 ride) was zig-zaging. I couldn't make that work. I just used my smallest gear. 30x25. I have a pretty standard road tripple.
At the San Francisco Grand Prix earlier this month, sections of the course were up to 15% grade (29.4 ft. climb over 196.6 ft, according to DeLorme Topo - but I think it might be more on the order of 11% or 12%. Isn't there some rule of thumb about how steep streets can be for cars?). Some of the guys chose to zig-zag it (obviously not bunched up in the peloton) and they climbed faster than the guys who went straight up the hill.

I've got one section of a ride in Seattle - 41.9 feet rise over 298.2 ft, which should calculate to a 295.2 ft actual distance? It's part of a bigger hill, but it's a killa.
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Old 09-23-05, 05:41 PM   #22
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Well, I just got one of these


so I'll doublecheck my Google Earth results.
I gather that thing tells you what the grade is at a particular spot? If so, it might not match up at all - or just match up by luck. I've never had a grade where the steepness stayed the same all the way up. The longer the hill, the more likely you have lots of changes. Which one you use for a measurement will determine your result.
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Old 09-23-05, 07:12 PM   #23
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Originally Posted by Fat Boy Biker
JavaMan will have to explain his formula. My formula is more accurate than the simple elevation/distance. You can see in my example the real answer is 10.05% (rounded) and the close answer was 10%. Believe me, .05% wont make any difference, but as Sir Scott said, the steeper the hill, the more inaccurate it becomes. But trust me, if you can climb a 20% grade, you are the man in my books.

Steve
-isn't math fun

P.S. My formula calculates the real horizontal lentgh using c squared = a squared + b squared, where the distance up the hill is c and the elevation gain is a, and the real horizontal distance is b. Then I used the elevation / real horizontal distance to get the "real" grade.
The horizontal length you calculated using Pythagorus is not necessary to find explicitly. The inverse sin or ASIN of (elevation gain/distance up the hill) gives you the angle of the hill. Since you now know the angle, you just take the TAN of that angle to get the grade. Then multiply by 100 for percent. That's the formula. RISE is in feet, DISTANCE is in miles.

%GRADE = TAN(ASIN(RISE/(5280*DISTANCE)))*100
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Old 09-23-05, 07:40 PM   #24
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If anyone's interested, I've included a "Climbing Calculator" in my CycliStats training diary program. It calculates the grade of any hill, and allows you to estimate how much time the climb will take. It also calculates your "rate of climb" (in feet or meters climbed per hour).



And, yes, I think it's a pretty neat tool, especially if you ride much in the mountains.

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Old 09-23-05, 08:56 PM   #25
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MAC friendly?
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