Testing ranks rather than actual numbers with parameteric tests An answer on this page Non parametric analysis  states: 

"You can perform ANOVA using ranks of your variable instead of
  variable itself. This will lead to the same results with
  Kruskal-Wallis test."

Suppose I have following vector and I convert it to ranks as follows:
vector:  30,45,21,11,27,26

ranks:  5 6 2 1 4 3

Do following tests make any sense?
1. Pearson correlation between ranks of 2 vectors.

2. paired t.test of ranks of 2 vectors of same length

3. unpaired t.test of ranks of 2 vectors of differing length

Will they be like non-parametric tests? Thanks for your insight.
 A: Doing them out of order:
1. is called the Spearman correlation
3. I assume we're doing the equal-variance version of the test; similar comments would apply to a Welch-Satterthwaite-type test. If you rank the samples by ranking them combined together, then compute the t-statistic on those ranks, then a permutation test on that statistic is equivalent to a Wilcoxon-Mann-Whitney. It's possible to show the two are monotonically related and the permutation test equivalence follows immediately. Here's some simulated data  showing how they relate monotonically:

(each point represents a pair of samples of size 10 from normal distributions; there are thousands of samples shown, though most points are coincident with other such points)
2. If you rank the samples by ranking them combined together, then compute the paired t-statistic on those ranks, then a permutation test based on that statistic won't be equivalent to a Wilcoxon signed rank test, but is correlated with it. It will still be nonparametric; I haven't made any attempt to see how well it performs in practice; I expect it will have lower power at the normal. (If you have some other way of ranking, or of organizing the test, you'll need to be more specific.)

As to the question of ranking them separately:
3. Let the first sample have $n$ observations and the second sample have $m$ observations, drawn from a continuous distribution. The first sample will consist of the ranks $1, 2, ..., n$ and the second sample will consist of the ranks $1, 2, ..., m$, and so the test statistic will always be exactly the same, no matter what values the sample consist of. 
For example, if the first sample has 5 values in it and the second sample has 4 values in it, all changing the sample can do is change the order of the ranks. The test statistic will always be 0.5092. If the two sample sizes are the same the test statistic will always be 0.
This wouldn't make sense as a statistic.
2. both samples will consist of the ranks $1, 2, ..., n$; the numerator of the t-test will always be $0$; different samples can change the denominator, but the t-statistic will always either be $0$ or - when both samples have their ranks in the same order - undefined (0/0).
This wouldn't make sense as a statistic.
