You also cannot equate a linear size increase with a linear speed increase, since the circumference of the disk increases quadratically. With respect to cache saturation and real-world testing, most users find a 35% variation in transfer speed from the outer to the inner surface of the disk platter (e.g. 100 MB/s to 65 MB/s). See the bottom of this answer for sample calculations.
While generally drives do write from the outside of the platter to the inner portion, data fragmentation often skews data across the platter. With a multi-platter drive, this situation becomes increasingly complex, as data may be distributed across more then one platter.
I have heard the "don't fill your hard drive" quote in the past, but unless you are dealing with a system disk (e.g. operating system or swap/page file), you can ignore this issue. Do note that you may have troubles defragmenting a disk without enough free space.
It is worth noting that you should not fill a solid-state drive to it's maximum capacity.
To prove my speed formula, let us assume that 40% of the platter's area is taken up by the spindle, and let us also assume that the platter is 3.5" in diameter, so we have an inner radius of 1.75". That means that the inner radius of the platter is given as 1.75 times the square root of 0.4, or about 1.11" (remember, area = pi * radius ^ 2, so work backwards).
Then, we compute the inner and outer circumferences as just C = 2 * pi * radius, yielding an outer circumference of ~11" and an inner circumference of 6.95". Since the circumference dictates the linear velocity, we see that the inner circumference will have a linear velocity of only 63.2% of the outer circumference - or in other words, 36.8% slower.
If you work out the math, you can prove that the speed decrease from the outer edge to inner edge of the platter is given by 1 minus the square root of the spindle size proportion (e.g. 1 minus the square root of 0.4 in our case, which yields 1 - 0.632 = 0.368).