The previous post Run Faster: Shoe Weight and Performance summarizes the conclusions of this post in a manner accessible to all readers. In short a lightweight flat was found to improve 5k race times by 1.4% when compared to a full cushioned trainer. The returns diminished for 10k and ½ marathon distances with improvements of 1.2 and 1.1% respectively. This post is written for readers interested in the supporting mathematical analysis, which employs basic physics and a study of the Bio mechanics of running. Please feel free to refute any assumptions or calculations. I am happy to discuss.

A visual examination of various runners and their form was necessary to create a proper model. Using a side profile and a stationary video camera is the best way to capture form and can be done at home on a treadmill or a sidewalk. In runners with good form over striding is absent. Cadence or foot turnover is a constant 180 foot strikes per minute. The trailing foot rises to the height of the knee. This same model does not hold true for runners who shuffle at lower speeds. This link contains good video examples of various running form midway down the page.

With respect to shoe weight no difference was found in the overall path of travel for forefoot vs. heel striking, thus the energy cost was assumed to be the same. Some advocates of barefoot running compare the best of fore foot strikes to the worst of over striding heel strikes and are intellectually dishonest in this respect. While heel striking may encourage over striding, over striding is a separate issue.

During a stride the trailing foot is accelerated forward and becomes the lead foot. The foot is then accelerated upward and becomes the trailing foot once again as the other leg strides forward. Two distinct components exist, vertical and horizontal. From basic physics the Energy required to accelerate a mass a certain distance can be found by:

E = F∆d

Where

**E**is the energy in Joules,**F**is the force in Newton’s, and**∆d**is the distance of each stride in meters. The force can be found by:F = ma

Where

**m**is the mass in kilograms and**a**is the acceleration in m/s^{2}. The mass represents the weight of one shoe in this study. The shoes selected for comparison were the Nike Free 3.0 V2 and Nike Voomero. These shoes were selected as one represents a minimalist shoe and the other a full cushioned traditional trainer. Weight can be converted from ounces to grams to kilograms by way of the conversion factors:1 oz = 28.348 g

1 Kg = 1000 g

m

_{1}= 6.7oz = 0.190 Kg (Nike Free)m

_{2}= 10.4oz = 0.295 Kg (Nike Voomero)To find the vertical component of the energy

**E**required to accelerate one shoe is fairly simple. Acceleration is set equal to gravity, 9.8 m/s_{V}^{2}. The distance traveled was found to be the height of the knee for all paces. For me this distance is 0.457 m. These values are substituted back into the Force and Energy equations to find the energy of the vertical component.a

_{V}= gravity∆d

_{V}= Knee HeightThe horizontal component of the energy

**E**required to accelerate one shoe requires additional consideration to first find the acceleration and distance. Unlike the vertical component, the horizontal component is dependent on pace. The horizontal change in distance is found by:_{H}∆d

_{H}= (1608 m)/(pace • cadence)Where pace is minutes per mile and cadence was earlier mentioned as 180 foot strikes per minute. There are 1608 meters in 1 mile. This calculation will give the distance of each stride in meters and varies according to pace. Faster paces have longer strides than slower paces. To find the acceleration of the horizontal component requires application of an equation of motion. This is the same math behind foot pods that calculate speed and distance using an accelerometer and time.

∆d

_{ }= ½ (a)( t^{2})Which can be rewritten as:

a

_{H}= 2∆d_{H}/t^{2}Where

**t**is the time in seconds of one stride and is found by dividing cadence by 60 seconds. The time is a constant 0.333 seconds for all paces. Horizontal acceleration will be greater for faster paces.t= Cadence / 60s = 0.333 s

After solving for horizontal distance and acceleration these values are substituted back into the Force and Energy equations to find the energy of the horizontal component. Once the Energy of the horizontal and vertical components are calculated they can be neatly summed together and converted in food calories. Pace and race distance should be expressed in miles.

E

_{TOT}= (E_{H}+ E_{V})( pace • cadence • race distance)/4184Where

**E**is the total energy required to accelerate the shoe in food calories for the calculated race distance._{TOT}**E**is the horizontal energy component in Joules and_{H}**E**is the vertical energy component in Joules. The conversion factor of 1 C = 4184 J is used to convert Joules into food calories. Once the total energy required to run with a shoe a fixed race distance is found and expressed in food calories the energy can be compared to empirical calorie data from runs. The energy cost as a fraction of the calories burned per race distance can be found by:_{V}η = E

_{TOT}/ ((Cal_{AVG }- Cal_{BMR}) • Race Distance)Where

**Cal**_{AVG}_{ }is the average calorie burn per mile taken from empirical data. This value will increase for longer distances as a function of cardiac drift.**CAL**represents calories burned by way of BMR per mile and is subtracted from the empirical value CAL_{BMR}_{AVG}to isolate the additional energy expended during running. My personal BMR of 1800 C was used for this study. This fraction can be multiplied by time and distance to find the associated time penalty**t**in seconds incurred by shoe mass:_{p}CAL

_{BMR}= BMR /24 hours / 60 minutes • Pacet

_{P}= η • pace • race distanceThe tables below summarize the results of the calculations:

Race Distance [mi] | Pace [min/mi] | ∆d _{H} [m] | a _{H} [m/s^{2}] | ∆d _{V} [m] | a _{V} [m/s^{2}] | |

5k (3.1) | 6:00 | 1.49 | 26.84 | 0.46 | 9.8 | |

10k (6.2) | 6:20 | 1.41 | 25.43 | 0.46 | 9.8 | |

½ mar (13.1) | 6:44 | 1.33 | 23.93 | 0.46 | 9.8 |

Race Distance [mi] | Cal _{AVG} [C/mi] | Cal _{BMR }[C/mi] | E _{TOT} [C]_{ } | |

m _{1} | m _{2} | |||

5k (3.1) | 97.8 | 7.5 | 6.8 | 10.5 |

10k (6.2) | 102.0 | 7.9 | 13.0 | 20.1 |

½ mar (13.1) | 108.2 | 8.4 | 26.2 | 40.7 |

Race Distance [mi] | η | t _{p} [s] | ||||

m _{1} | m _{2} | ∆ | m _{1} | m _{2} | ∆ | |

5k (3.1) | 0.024 | 0.038 | 0.014 | 27 | 42 | 15 |

10k (6.2) | 0.022 | 0.034 | 0.012 | 52 | 80 | 28 |

½ mar (13.1) | 0.02 | 0.031 | 0.011 | 106 | 165 | 59 |

Do these calculations make sense? Yes. The 1985 study Effects of shoes and foot orthotics on VO2 and selected frontal plane knee kinematics stated that oxygen consumption during running increased as the amount of mass they added to the foot increased; shoes and orthotics representing 1% of body mass increased oxygen consumption by 3.1%. Using this rule of thumb and my own 140lb body weight and assuming the 3.1% increase in VO2 correlates 1:1 to the energy calculations in this post:

A pair of Nike Voomeros represents 0.93% of total bodyweight, scaling VO2 yields a 2.9% increase in the energy required. A pair of Nike Free 3.0 V2 represents 0.6% of total bodyweight, scaling VO2 yields a 1.9% increase in the energy required. These values are close to the ½ marathon value calculated for η. The difference in value could be explained by the dampening effect of the cushioning which was not considered in this post. Shoe weight was the only factor in this analysis.

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