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| Acmlm's Board - I3 Archive - General Chat - "Road Gradients/Steepnesses or whatever" |
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Deleted User Banned                Since: 05-08-06 Last post: None Last view: 6297 days |
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| Yeah, it's another one of those questions. Anyway, I was just wondering, just before you get to a hill (this is in the UK by the way), there are these signs that say either "1 in 10" or "1/10" or as a percentage, e.g. "10%", but what exactly does that mean? There's a road in Caterham, Surrey (England), called Church Hill, which has '16%' on it. First I thought it was 16% of 360-degrees, but that's about 57-degrees, and I thought that's pretty steep, and then I found this website which has a list of 'Road Records', and said that a road in the UK had a '1 in 2.7-2.9' gradient, which in that case would mean it's upside down, which then led me to thinking it must mean '10%, 16% or what ever percent of 180-degrees', so a '10% gradient' would be on an incline of 18-degrees, and a '50% gradient' would be on an incline of 90-degrees, i.e. vertical. However, on this website, it had all these weird things about '1/10 on rail means sin(1/10) but on a road way it's tan(1/10), and things like '1/2.93 on a road is tan(arcsin(1/10)) and other weird trigonometrical mumbo-jumbo.
So, anyway, basically, what does '1-in-6' '10%' etc mean? (Or even, is there a formula that converts the 'percentage' to the 'degrees on incline' or whatever? |
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Prince Kassad   320 As you wish.        Since: 06-30-06 From: nowhere Last post: 6298 days Last view: 6298 days |
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| In most states '10%' means that if you drive 100 meters forward, you will go down 10 height meters. It's probably the same for the UK. | |||
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Deleted User Banned Since: 05-08-06 Last post: None Last view: 6297 days |
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| Well that would explain the trigonometry. So in other words, to work out the gradient in degrees, for a '10%', the 'opposite' and 'adjacent' sides are used, so it's 'Tan(gent)' we should use. The gradient = Tan(Opposite/Adjacent), in this case, Tan(10/100), which is.... less than 1?? Unless it should be Tan-1 (inverse tangent)?
EDIT: OK, I think I found a formula to convert the gradient as a percentage into degrees: =180/PI()*ATAN(0.1/100) <--- this is to calculate a '10%' gradient The general formula would be "=180/PI()*ATAN("slope"/100)" but it doesn't seem to work properly. Apparently a '100%' gradient is 45 degrees, but if I type that as a formula in Excel, it comes out as 0.572938698 radians*, or, in degrees, 32.8269693. In words the formula is "180 over pi, multiplied by the arctangent of 1 percent of the slope" in other words, a 15% gradient would be "180 over pi, multiplied by the arctangent of 1 percent of 15 percent, or "180 over pi, multiplied by the arctangent of 0.0015" but it comes out wrong. How would I go about putting the formula into excel? *Note: Excel seems to display the formula in 'radians' rather than 'degrees', so the the formula actually is "=DEGRESS(180/PI()*ATAN("slope"/100) EDIT 2: Don't worry, I just found another formula which does work: It's simply the arctangent of the slope, full stop. In excel, this is: =DEGREES(ATAN(A2)) I drew a flat line in Word, and then, changed its properties to change the angle of rotation to that of what that above formula gave as a '16% gradient' in degrees (just over 9.09 degrees). I've been on a road with a "16%" sign, but this line looks less steep than a road I went on with a "10%" gradient. The two roads in question, if you want to know are: Church Hill, Caterham, Surrey (UK) = 16% and Redstone Hill, Redhill, Surrey (UK) = 10%. (edited by TheUltimateKoopa #2 on 09-09-06 11:02 AM) (edited by TheUltimateKoopa #2 on 09-09-06 11:15 AM) (edited by TheUltimateKoopa #2 on 09-09-06 11:22 AM) |
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