I am trying to understand a Spice definition of a circuit and transient analysis on it. I have kind of "inherited" this and can't ask the original creator; my Spice memories from university are pretty rusty, and online manuals have not been of much help.

What I don't understand is the first segment here (snippet of whole file):

.define c1 6926.0 6233.4 7618.6
.define r1 0.16661 0.149949 0.183271
....
il   0 98 dc 0 ac 0 pwl (0s 1720.0a ... 6233s 1718.0a 6925s 1728.0a 7618s 1718.0a ....)
...
c01  1 0 [c1] ic= 89.974
r01  1 2 [r1]


In the third segment, I understand that capacitor c01 is defined as lying between nodes 1 and 0 and has initial conditions as given; resistor r01 is situated between nodes 1 and 2.
In the second segment, a piecewise linear current source is defined, with ampere values associated with timestamps.

What does the first segment define? The values next to the capacitor seem to be the timestamps from the current source, but why are they used for defining the capacitor/resistor?

I can provide other details as needed, but didn't want to clutter the post too much with useless information.

EDIT: The suggestion given below in comments is good, but my original question still stands. Is this not standard Spice, since no manual mentions it?


Question: Cubing a simple thing
Answer: There is a more intuitive way to do this that works like the FOIL method.

Write $(x+2)^3$ as $(x+2)(x+2)(x+2)$.

When we multiply these we look at all the ways we can pick one element from each of the $(x+2)$s, multiply each possibility and add them all up.

What do these look like? Well, from each $(x+2)$ you can pick either an $x$ or a $2$. Since we pick one from each, we will get three things multiplied together.

If we pick an $x$ from each we will get an $x\cdot x\cdot x$ which equals $x^3$. We can only do this one way so we get $x^3$. 

What if we pick an $x$ from only two of the $(x+2)$s? Then we have to pick a $2$ from the third. How many ways are there to do this? Well, we can pick an $x$ from the first two and a number from the third; we can pick an $x$ from the first and the third and a number from the second or we can pick an $x$ from the second and the third and a number from the first. This gives us $3(2x^2)=6x^2$. So far we have $x^3+6x^2$.

What if we pick only one $x$ and a $2$ from the other two? What do we get and how many ways are there to do this? We get $2\cdot 2\cdot x=4x$ and there are three ways to do this because we can pick the $x$ from either of the three. So we get $3(4x)=12x$ which gives us $x^3+6x^2+12x$ all together.

Are there any other possibilities? Yes, we can pick a number from each and there is only one way to do that so we get $2\cdot 2\cdot 2=8$ and finally we arrive at $(x+2)^3=x^3+6x^2+12x+8$.

This approach is easier than it seems if you practice a few and it works for $(x+a)^n$ for any positive n.