# 9.6.2 Ride Indexing Explained: Examples using SI units

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
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.

----------------
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.

----------------
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

----------------
Example 3:

How many Calories have I burned on my loop ride?

TC = index * K

TC = 145 * 23 = 3340

----------------
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

----------------
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.

----------------
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

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