High-capacity tape formats use helical scanning. Here I look at why that gives a higher capacity than linear recording with four tracks (one stereo track on each side) like in a compact cassette (C-cassette).
Figure 1. A simplified illustration of the tracks on a four-track tape (left) and helical scanning tape (right). In reality the spacing of the tracks is not as dense, to prevent leakage between the channels.
It's not about the speed at which the recording head travels. It is possible to slow things down and record the same spatial wavelengths, only it would take a longer time to do that or to read the recording later. Recording faster does not increase tape capacity. What it does is that it gives a higher data rate which is important for practical applications.
For information density, it doesn't matter what the angle of the tracks on the tape is. So for a simplified analysis we can rotate the helical scanning tracks to be horizontal, and we can look at just a 1/4 of the tape and consider that a single-track tape in case of the C-cassette:
Figure 2. One-track linear recording tape (left) and an n-track linear recording tape (right).
Let's write a spatial version of the Shannon–Hartley theorem that suits our simplified analysis:
$$C = nB \log_2 \left( 1+\frac{S}{nN} \right)\tag{1}$$
Here $C$ is the total capacity (bits/m of tape), $n$ is the number of tracks, $B$ is the spatial bandwidth (1/m) and $S/N$ is the reference signal-to-noise ratio, the ratio between signal power and noise power in the reference case that there is just a single track. If we increase from the reference case the number of tracks by a factor $n$, the signal power gets divided by a factor $n^2$ (signal in-phase between the tracks) and the noise power gets divided by a factor $n$ (independent noises in the tracks), hence the effective signal-to-noise ratio $\frac{S}{nN}$ in the formula. At the same time, the total capacity is the sum of the individual track capacities, hence the factor $n$ in front of the formula.
If we plot the capacity calculated by Eq. 1 as function of the number of channels, it would appear that we can increase the capacity indefinitely by just increasing the density of the tracks:
Figure 3. Proportional channel capacity as function of the number of tracks, for an out-of-the-hat signal-to-noise ratio of 50 dB in case of a single track.
The main thing that is wrong with this analysis is that the size of the magnetic grains is not truthfully infinitesimally small, so there will be all kinds of unaccounted for trouble (correlated noise between tracks) if the tracks become too narrow and densely spaced. But I would think that the result still holds up to some limit, that more and narrower tracks can store more information even when each has a lower signal-to-noise ratio than would a single, or four, tracks. So it is not just that we wouldn't have come up with the right modulation scheme for C-cassettes. The design of data tape formats with helical scanning really is superior in terms of information capacity expressed per tape length, for the same width of tape.
That is, unless I got something wrong in the analysis. :)