9.6.1 Ride Indexing Explained: Examples using English units
Description
This article is from the Misc Bicycles FAQs, by various authors.
9.6.1 Ride Indexing Explained: Examples using English units
Finding the cyclist's divisor, D:
First, calculate K.
I ride with an average flat-land speed of v=17 mph. My combined bike and body
weight is approximately 210 lbs., and I spend most of my time riding on the
hoods or on top, so I use a quadratic coefficient of air resistance a=0.24.
bike_power session:
% bike_power -wm 42 -wc 168 -v 17 -a 0.24
grade of hill = 0.0% headwind = 0.0 mph
weight: cyclist 168.0 + machine 42.0 = total 210.0 lb
rolling friction coeff = 0.0060 BM rate = 1.40 W/kg
air resistance coeff = (0.2400, 0)
efficiency: transmission = 95.0% human = 24.9%
mph F_lb P_a P_r P_g P_t P hp heat BM C Cal/hr
17.0 4.4 105 43 0 8 156 0.21 470 107 732 629
K = (629 Cal/hr) / (17 mi/hr) = 37 Cal/mi
Now calculate D.
D = (37 Cal/mi) / [(1.37E-3 Cal/(ft*lb))*(210 lbs)] = 129 ft/mi
This means that for me, I use as much energy to lift me and my bike 129
vertical feet as I use to pedal one mile on flat, windless ground at 17 mph.
I have decided to use F = 0.9 since I sometimes pedal on downhills, but not
always. So my indexing formula is:
index = d + n/129 + (g - n)/232
For the examples below, D will be accurate to +/- 1. In practice, given
that indexing is an approximation, one can round the figures to the nearest 10
or even to the nearest 50 without too much loss of accuracy. One might use
D = 150, and 2*F*D = 250 for in-the-head calculations.
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Example 1:
Suppose I take a 75 mile loop ride with 4000 feet of climbing. What is my
index?
For a loop ride, I use the following formula:
index = d + g/232
The index for this ride is:
index = 75 + 4000/232 = 92
This means I used the same amount of energy on this ride as I would use to
ride 92 miles on flat ground at my normal riding pace.
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Example 2:
Suppose I rode my loop in 6 hours total time (including lunch and rests) and
my Avocet 50 shows an average speed of 17.2 mph. Calculate the index rate of
progress (irp), and the moving index rate of progress (mirp).
irp = 92/6 = 15 mph
mirp = 92 / (75 / 17.2) = 21 mph
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Example 3:
How many Calories have I burned on my loop ride?
TC = index * K
TC = 92 * 37 = 3404
That's almost a pound of fat!
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Example 4:
Suppose I start the ride at sea level and end at 1300 feet. What is my index?
Since the ride starts and ends at a different elevation, we must use the
complete formula.
index = d + n/129 + (g - n)/232
index = 75 + 1300/129 + (4000-1300)/232 = 96
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Example 5:
Suppose I've ridden 20 miles with 2500 feet of climbing in 2 hours. How
long will it take me to ride the remaining 55 miles and 1500 feet of climbing
at the same pace? (Assume that I am currently at my starting elevation.)
My current index is 20 + 2500/232 = 31.
My index average speed is (31 miles) / (2.0 hours) = 15.5 mph
The index of the remaining ride is: 55 + 1500/232 = 61 miles.
At my current index speed, I should be able to complete the ride in
(61 miles) / (15.5 mph) = 3.9 hours or 3:56.
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Example 6:
I wish to compare the 100 mile and 200 kilometer routes for the 1992 Sequoia
Century: Both rides start and end in the same place, so we can use the
simplified formula.
100 mi route: d = 100, n = 0, g = 10,800 200 km route: d = 121, n = 0, g = 9,800
The index for the 100 mi route is: 100 + 10800/232 = 147 miles
The index for the 200 km route is: 121 + 9800/232 = 163 miles
Number of Calories burned:
100 mi route: K * 147 = 37 * 147 = 5440 200 km route: K * 163 = 37 * 163 = 6030
One might conclude that the 200 km route was more difficult. One should be
aware that difficulty or pain is subjective, and the formulas given above
take into account neither steepness of grade nor availability of efficient
gearing for a particular grade.
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