Load-Exertion Tables And Their Use For Planning – Part 4

Load-Exertion Tables And Their Use For Planning – Part 4

Previous Part:

Load-Exertion Tables And Their Use For Planning – Part 3

Epley, or Not to Epley. That Is the Question

In the previous part, we have used a set of Epley’s equations to estimate both group/pooled and individual reps-max profiles, as well as to create progression tables (Equations 1). One characteristic of Epley’s equation is that 100% 1RM is achieved with 0RM, which is not an ecologically valid assumption. What the hell is 0RM in the first place? What we want is that 100% 1RM is associated with 1RM.

\[\begin{equation}
\begin{split}
nRM &= \frac{1 – \%1RM}{k \times \%1RM} \\
\%1RM &= \frac{1}{k \times nRM + 1}
\end{split}
\end{equation}\]

Equations 1

It is easy to check this feature by using Equations 2:

\[\begin{equation}
\begin{split}
nRM &= \frac{1 – \%1RM}{k \times \%1RM} \\
nRM &= \frac{1 – 1}{k \times 1} \\
nRM &= \frac{0}{k} \\
nRM &= 0 \\
\\
\%1RM &= \frac{1}{k \times nRM + 1} \\
\%1RM &= \frac{1}{k \times 0 + 1} \\
\%1RM &= \frac{1}{1} \\
\%1RM &= 1
\end{split}
\end{equation}\]

Equations 2

I am not sure why they selected this model definition in the first place. This feature of the Epley’s model is very confusing, which will be quite obvious later in this article series when I will introduce novel models and estimate 1RM alongside with the k parameter.

Luckily, this can be easily solved by using, what I call, Modified Epley’s (set of) Equations (Equations 3). I have renamed the k parameter to kmod to avoid confusion and to prevent using them interchangeably.

\[\begin{equation}
\begin{split}
nRM &= \frac{(kmod – 1) \times \%1RM + 1}{kmod \times \%1RM} \\
\%1RM &= \frac{1}{kmod \times (nRM – 1) + 1}
\end{split}
\end{equation}\]

Equations 3

Using this set of equations (Equations 3) correctly assumes that 1RM is achieved at 100% 1RM. I will leave you to check it out the same way we have done in Equation 2.

To demonstrate the difference between the two model definitions, I will again create a group/pooled model using our 12 athletes. The results are depicted in Figure 1. Estimated k is equal to 0.032, and estimated kmod is equal to 0.036.

Figure 1: Group or pooled model fit using Epley (Equations 1) and Modified Epley (Equations 3) model definitions. This approach uses all the athletes to estimate the “average” profiles

Model performance is shown on Table 1 using five estimators: (1) mean-absolute-error (MAE), (2) root-mean-squared-error (RMSE), (3) maximal error (maxErr), (4) error interquartile range (IQR), and (5) variance explained (\(R^2\)) 4 . Since we are dealing with Group/pooled models, both models performed poorly. Please note that I have used Epley’s model to generated the data in the first place.

Model MAE (Reps) RMSE (Reps) maxErr (Reps) IQR (Reps) R2
Epley 2.30 3.02 -7.82 3.76 0.63
Modified Epley 2.34 3.04 -8.14 3.62 0.63

Table 1 – Group/Pooled model performance on the training data set. MAE = mean-absolute error; RMSE = root-mean-squared-error; maxErr = maximal error; IQR = error interquartile range; R2 = variance explained

Other model definitions, besides Epley and Modified Epley, can be used to model (or map out) max reps and %1RM relationship. The most common model used in data science is a simple linear regression (Equation 4).

\[\begin{equation}
nRM = slope \times \%1RM + intercept
\
\end{equation}\]

Equation 4

Simple linear regression is, simply put, a line. To estimate the best fit line through our observed data, we need to estimate two parameters: slope and intercept (Equation 4). This implies that we need at least three observations. When it comes to individual models, this implies that one needs to do three sets to fail. With Epley and Modified Epley model, one needs to do only two sets to failure. This is a practically meaningful difference.

Regression’s bigger brother, polynomial regression, particularly 2nd degree polynomial regression is also often utilized in modeling max reps and %1RM relationship (Equation 5).

\[\begin{equation}
nRM = slope_1 \times \%1RM + slope_2 \times \%1RM^2 + intercept
\end{equation}\]

Equation 5

Second degree polynomial have three parameters to be estimated: slope 1, slope 2 and intercept. The problem with this, particularly with the individual model, is that one needs at least four observations (i.e., sets to failure) for the parameters to be estimated. The additional problem involves different nRM associated with 100% 1RM (similar to the original Epley’s model), and for the polynomial model, we might have very weird predictions outside of observations range (i.e., predicting for 60% 1RM, when we have 70-90% observational data). Figure 2 depicts simple linear regression and 2nd-degree polynomial fit models. Note how the models behave around 100% 1RM.

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I am a physical preparation coach from Belgrade, Serbia, grew up in Pula, Croatia (which I consider my home town). I was involved in physical preparation of professional, amateur and recreational athletes of various ages in sports such as basketball, soccer, volleyball, martial arts and tennis. Read More »
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