This article is from the Misc Bicycles FAQs, by various authors.

Finding the cyclist's divisor, D:

First, calculate K.

I ride with an average flat-land speed of v=27 kph. My combined bike and body

weight is approximately 95 kg, 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:

% bp +M -wm 19 -wc 76 -v 27 -a 0.24 -M grade of hill = 0.0% headwind = 0.0 mph weight: cyclist 167.5 + machine 41.9 = total 209.4 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 16.8 4.3 101 42 0 8 151 0.20 455 106 712 612

K = (612 Cal/hr) / (27 km/hr) = 23 Cal/km

Now determine D.

D = (23 Cal/km) / [(9.91E-3 Cal/(m*kg))*(95 kg)] = 24 m/km

This means that for me, I use as much energy to lift me and my bike 24

vertical meters as I use to pedal one kilometer on flat, windless ground at 27

kph.

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/24 + (g - n)/44

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 5

or even to the nearest 10 without too much loss of accuracy. One might use

D = 25, and 2*F*D = 50 for quick, in-the-head calculations.

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Example 1:

Suppose I take a 100 kilometer loop ride with 2000 meters of climbing. What is

my index?

For a loop ride, I use the following formula:

index = d + g/44

The index for this ride is:

index = 100 + 2000/44 = 145

This means I used the same amount of energy on this ride as I would use to

ride 145 km on flat ground at my normal riding pace.

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Example 2:

Suppose I rode the loop in 6 hours total time (including lunch and rests) and

my Avocet 50 shows an average speed of 23.0 kph. Calculate the index rate of

progress (irp), and the moving index rate of progress (mirp).

irp = 145/6 = 24 kph

mirp = 145 / (100 / 23.0) = 33 kph

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Example 3:

How many Calories have I burned on my loop ride?

TC = index * K

TC = 145 * 23 = 3340

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Example 4:

Suppose I start the ride at sea level and end at 400 meters. What is my index?

Since the ride starts and ends at a different elevation, I must use the complet

e

formula.

index = d + n/24 + (g - n)/44

index = 100 + 400/24 + (2000-400)/44 = 153

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Example 5:

Suppose I've ridden 50 km with 500m of climbing in 2 hours. How long will it

take me to ride the remaining 70km and 1500m of climbing at the same pace?

(Assume that I am currently at my starting elevation.)

My current index is 50 + 500/44 = 61.

My index average speed is (61 km) / (2 hours) = 31 kph.

The index of the remaining ride is: 70 + 1500/44 = 104 km.

At my current index speed, I should be able to complete the ride in

(104 km) / (31 kph) = 3.35 hours or 3:21.

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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 = 161, n = 0, g = 3300 200 km route: d = 195, n = 0, g = 3000

The index for the 100 mi route is: 161 + 3300/44 = 236 km The index for the 200 km route is: 195 + 3000/44 = 263 km

Number of Calories burned:

100 mi route: K * 236 = 23 * 236 = 5430 200 km route: K * 263 = 23 * 263 = 6050

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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