Thanks, Alex, for looking into this.

It is important to remember that contrasts are simply a formalization of your null hypothesis, that is, given your specific hypothesis about the mean

differences under the null hypothesis, you can derive the appropriate contrast.

I try to give an intuitive and simplified derivation of equal and unequally spaced contrasts for testing a linear trend.

DISCLAIMER: this is a loosely formalized derivation.

Let's first look at the derivation of a linear trend contrast with equal spacing.

Assume, we have four time points of measurement. First, we state that the mean difference between two successive time points is the same,

namely \Delta_t:

\mu_2 = \mu_1 + \Delta_t

\mu_3 = \mu_2 + \Delta_t

\mu_4 = \mu_3 + \Delta_t

A valid contrast must sum up to one, therefore,

we require:

\sum{\mu_i} = 0

The above are four equations with five unknowns,

so we have to choose one arbitrary additional constraint.

Let's fix the time to one time unit between measurements:

\Delta_t = 1

Now, we have obtained a set of linear equations that can be

easily solved (and for the sake of brevity, I will not paste it here):

\mu_1 = -1.5

\mu_2 = -0.5

\mu_3 = +0.5

\mu_4 = +1.5

This is a contrast for a linear trend. Now, if you know the exact time

points of measurement, e.g., measurements at T_0 = 0, T_1 = 6, T_2 = 8, T_3 = 10,

you can derive individual \Delta_{ti} by taking the difference scores of

each two successive measurements and solve the same set of equations

as above. In this example, we would obtain:

\Delta_{t1} = T_1 - T_0 = 6

\Delta_{t2} = T_2 - T_1 = 2

\Delta_{t3} = T_3 - T_2 = 2

\mu_2 = \mu_1 + \Delta_{t1}

\mu_3 = \mu_2 + \Delta_{t2}

\mu_4 = \mu_3 + \Delta_{t3}

Solving this under the same constraints as above, we obtain the following contrast:

\mu_1 = -6/24

\mu_2 = 0

\mu_3 = 2/24

\mu_4 = 4/24

Of course, the above can be similarly calculated for other polynomial contrasts and other number of time points.