For a problem this small the slopes are easily computed with a simple raster calculation. Given that the years are consecutive, let's name the rasters [y.1], [y.2], [y.3], [y.4], and [y.5] in temporal order. The slope grid is
(2/10) * ([y.5] - [y.1]) + (1/10) * ([y.4] - [y.2])
For other than five rasters--but still assuming they represent consecutive times--there is a similar formula. Each raster [y.i], for i = 1, 2, ..., through n, gets multiplied by a coefficient and all these results are added up. The coefficients are obtained by writing down the numbers
12, 24, 36, ..., 12n
and subtracting 6(n+1) from them. For instance, with n=8 we would subtract 6(8+1) = 54 from each, giving the eight numbers
-42, -30, -18, -6, 6, 18, 30, 42
These would multiply the rasters in temporal order. It's convenient to pair them by common coefficient sizes so you could write this out as
42 * ([y.8] - [y.1]) + 30 * ([y.7] - [y.2]) + 18 * ([y.6] - [y.3]) + 6 * ([y.5] - [y.4])
That reduces the amount of writing and the number of grid multiplications that are done. Finally, divide the result by n^3 - n. In the case n = 8, n^3 - n = 512 - 8 = 504. The net effect (if you want to compare this to other formulas) would be to multiply the input rasters by the coefficients
-1/12, -5/84, -1/28, -1/84, 1/84, 1/28, 5/84, 1/12
and add up the results.
In more general situations, where there may be varying intervals between the rasters, there is still a similar formula: the slope grid is always a linear combination of the rasters, but the coefficients will be less regular. The coefficients can be found from the general formula (X'X)^(-1)X' where X is the n by 2 "design matrix" having a column of n 1's and a second column set to the times of the grids.
Usually the wrong way to do this is to loop over all the cells, pick out the cell values in the n rasters, and send them (and the times) to a line-fitting routine. That's much more work than necessary, because in effect each call to that routine is working out the same coefficients millions of times over (once for each cell). If, however, the rasters have a large number of missing values occurring in many different patterns, this longer way would make sense, for then you could obtain slopes even where one or more of the rasters is missing a value but the remaining rasters do have values.
Best Answer
Just chatting about this and wondering whether you could approach the problem by treating the input grids as a binary stream. This would allow you to combine them to give a unique summary integer for the sequence - ie 01110101 = 117. This value could then be reclassified to give the maximum number of consecutive 1s.
Here's an example showing one way to combine eight grids:
Bitwise operations could also be pressed into service for this step. Alternatively, you can use combine followed by a field calculation. (The field calculation will have an expression similar to the preceding one.)
The reclassification table has to provide max run lengths for all values between 00000000B = 0 and 11111111B = 255. In order, here they are:
This approach is limited to about 20 grids in ArcGIS: using more than this can create an unwieldy attribute table. (
Combineis specifically limited to 20 grids.)