 # 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
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.

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

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

----------------
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!

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

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

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