Pace

Sounds simple, right? Pace is just speed over distance. But there is more to it than meets the eye.

What is the GAP?

GAP stands for Gradient-Adjusted Pace. In simple words, this pace calculation accounts uphill and downhill movement. The calculation differs depending on your activity type.

Key Differences Between Running and Cycling

1. Rolling Resistance Matters:
2. In running, energy is mostly used to overcome gravity and metabolic costs.

3. In cycling, rolling resistance plays a role, meaning the energy cost of different gradients isn’t as extreme as in running.

4. Aerodynamic Drag Becomes Significant:

5. At higher cycling speeds, air resistance (drag) is the dominant force on flat terrain, making speed adjustments different from running.

6. Power Output Defines Effort, Not Just Metabolic Cost:

7. Cycling effort is measured in watts (W), and power-to-speed relationships follow a cubic relationship on flat ground due to aerodynamic drag.

Running

Per linear correction

1. Gradient Conversion

A gradient $s$% corresponds to an angle $\theta$, which can be approximated as:

$\theta \approx \tan^{-1}(s/100)$

1. Speed Adjustment Formula

$v_{\text{flat}} = v_{\text{incline}} \times \left(1 + \frac{s}{100} \right)$

Example Calculation

Let’s say you’re running 12 km/h on a 6% incline ($s = 0.06$):

$v_{\text{flat}} = 12 \times \left(1 + \frac{6}{100} \right)$

$v_{\text{flat}} = 12 \times 1.06$

$v_{\text{flat}} = 12.72 \text{ km/h}$

If you’re running 12 km/h on a 6% incline, it’s equivalent to running 12.72 km/h on flat ground using the linear correction model. But there is more:

The Minetti Metabolic Cost Model

A widely accepted model for running energy cost was developed by Minetti et al. (2002). It estimates the extra energy required to run at different gradients:

$C(s) = 155.4s^5 - 30.4s^4 + 5.4s^3 + 0.68s^2 + 1.94s + 3.6$

Where:

• C(s) is the metabolic energy cost (Joules per kilogram per meter).
• s is the gradient as a decimal (e.g., a 5% incline → s = 0.05 ).

This formula accounts for the fact that running downhill beyond a certain slope (~ -10%) becomes inefficient due to braking forces. The baseline cost for running on flat terrain is 3.6 J/kg/m. To estimate equivalent flat pace, we can adjust speed using the metabolic cost ratio:

$v_{\text{flat}} = v_{\text{gradient}} \times \frac{C(0)}{C(s)}$

Where:

• $v_{\text{gradient}}$ is your actual speed on the slope.
• $v_{\text{flat}}$ is the estimated equivalent flat speed.
• $C(0) = 3.6$ is the energy cost on flat ground.
• $C(s)$ is the energy cost at the given gradient.

Example Calculation

Let’s say you’re running 12 km/h on a 6% incline ($s = 0.06$):

1. Compute the metabolic cost at 6% slope:

   $C(0) = 3.6 J/kg/m (metabolic cost on flat ground).$

   $C(0.06) = 5.14 J/kg/m.$

2. Compute adjusted speed:

   $v_{\text{flat}} = 12 \times \frac{5.14}{3.6}$

   $v_{\text{flat}} \approx 17.1 \text{ km/h}$

Conclusion:
If you’re running 12 km/h on a 6% incline, it’s equivalent to running 17.1 km/h on flat ground in terms of effort. That is quite a difference compared to the linear model with a corrected pace of 12.72 km/h.

Why Is the Difference So Big?

• The linear model assumes a fixed percentage effort increase per gradient, which is inaccurate for steep inclines.
• The Minetti model accounts for the true exponential increase in energy demand, meaning the difference in equivalent speed grows significantly as inclines get steeper.

For gradients below ~3%, the linear model is reasonably close, But for steeper inclines, the Minetti model is much more accurate.

Cycling

The required power output $P$ for cycling is influenced by:

• Gravitational resistance (on inclines)
• Rolling resistance (constant but varies slightly with slope)
• Aerodynamic drag (depends on speed and wind conditions)

Cycling Power-Based Gradient Adjustment

For a steady climb at constant power, the equivalent speed on flat ground is:

$v_{\text{flat}} = v_{\text{incline}} \times \left( 1 + \frac{s}{200} \right)$

For descending, the correction factor is smaller because aerodynamic drag limits downhill speed increases:

$v_{\text{flat}} = v_{\text{incline}} \times \frac{1}{1 - \frac{s}{100}}$

Example Calculation

Let’s say you’re cycling 20 km/h on a 6% incline ( s = 6 ).

$v_{\text{flat}} = 20 \times \left( 1 + \frac{6}{200} \right)$

$v_{\text{flat}} = 20 \times 1.03$

$v_{\text{flat}} \approx 20.6 \text{ km/h}$

Going downhill is not as "easy" to calculate. It increases your effective speed, but - due to air resistance - not linearly.

Forces Acting on a Cyclist on a Slope

The power required to maintain a certain speed while cycling downhill is influenced by:

• Gravity $F_g$ : Helps accelerate the cyclist downhill.
• Rolling Resistance $F_r$ : Always resists motion.
• Aerodynamic Drag $F_d$ : Increases quadratically with speed.

The total force equation on a downhill slope $( s )$ is:

$F_{\text{net}} = m g \sin(\theta) - F_r - F_d$

Where:

• $m$ = cyclist mass,
• $g$ = gravity (9.81 m/s²),
• $\theta = slope angle ( \theta \approx \tan^{-1}(s/100) )$,
• $F_r = C_r m g \cos(\theta)$ (rolling resistance, where $C_r$ is the rolling resistance coefficient),
• $F_d = \frac{1}{2} C_d A \rho v^2$ (aerodynamic drag, where $C_d$ is the drag coefficient, $A$ is frontal area, $\rho$ is air density, and $v$ is speed).

Nonlinear Speed Adjustment Formula

Using an approximate power-speed relationship (since power scales with v^3 on flat ground due to drag):

$v_{\text{flat}} = v_{\text{incline}} \times \sqrt[3]{1 + \frac{|s|}{100}}$

Where:

• $|s|$ is the absolute gradient in percent.

Example Calculation

Let’s say you are cycling at 40 km/h on a -10% downhill slope. Using the nonlinear formula:

$v_{\text{flat}} = 40 \times \sqrt[3]{1 + \frac{10}{100}}$

$v_{\text{flat}} = 40 \times \sqrt[3]{1.1}$

$v_{\text{flat}} \approx 40 \times 1.032$

$v_{\text{flat}} \approx 41.3 \text{ km/h}$

With the linear model, the estimated flat-equivalent speed would have been 44 km/h (too high). With the nonlinear model, it’s more realistic at 41.3 km/h, since aerodynamic drag limits speed gains.