accuracy - SetAccuracy behavior

I'm puzzled by the output I get from SetAccuracy. According to the documentation, when SetAccuracy is used to increase the accuracy of a number, the number is padded with zeros. But, let's take a look at a couple of examples:

SetAccuracy[1.2, 5]
(* 1.2000 *)
SetAccuracy[1., 5]
(* 1.0000 *)
SetAccuracy[0.2, 5]
(* 0.2000 *)

These examples seem to work properly, so why does it behave differently in this case?

SetAccuracy[0., 5]
(* 0.*10^-5 *)

What should I do to get a zero with four trailing zeros?

Update I'm asking this question, because I need to export data to a txt file and I would like to avoid having 0.*10^-5 sort of numbers.

Answer

The comments by image_doctor led me to the answer I was looking for:

StandardForm@NumberForm[1.2, {20, 4}, ExponentFunction -> (Null &)]

(* 1.2000 *)
StandardForm@NumberForm[1., {20, 4}, ExponentFunction -> (Null &)]
(* 1.0000 *)
StandardForm@NumberForm[0.2, {20, 4}, ExponentFunction -> (Null &)]
(* 0.2000 *)
StandardForm@NumberForm[0., {20, 4}, ExponentFunction -> (Null &)]
(* 0.0000 *)

The data are then consistent and can be easily exported.

Comments

plotting - Plot 4D data with color as 4th dimension

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plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

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