I've been surveying the nine examples of V-loops given by daj 
< here > because I've never been clear what distinguishes them from SK Loops. What they all have is an SK Loop with a common digit in the digit pairs used to link the two rows and two columns in the loop. This digit will never appear as a given in the 8 houses that contain the loop which provides a way to recognise when an SK Loop is also a V-loop.   
Example: 9.......1.3...4.7...6...2...5.3.2.......6........78.5...2...6...4.7...3.1.......9#8400;tax;tarek-ultra-0204
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-        *-----------------------*-----------------------*-----------------------*
 | <9>    278    4578    | 2568   2358   3567    | 3458   468    <1>     | (12)b2
 (58)r2 | 258    <3>    158     | 125689 12589  <4>     | 589    <7>    568     |
 | 4578   178    <6>     | 1589   13589  13579   | <2>    489    3458    | (69)b3
 *-----------------------*-----------------------*-----------------------*
 | 4678   <5>    14789   | <3>    149    <2>     | 14789  14689  4678    |
 | 23478  12789  134789  | 1459   <6>    159     | 134789 12489  23478   |
 | 2346   1269   1349    | 149    <7>    <8>     | 1349   <5>    2346    |
 *-----------------------*-----------------------*-----------------------*
 | 3578   789    <2>     | 14589  134589 1359    | <6>    148    4578    | (69)b7
 (58)r8 | 568    <4>    589     | <7>    12589  1569    | 158    <3>    258     |
 | <1>    678    3578    | 24568  23458  356     | 4578   248    <9>     | (12)b9
 *-----------------------*-----------------------*-----------------------*
 (78)c2                                           (48)c8
Alongside the grid the linking digits used in the SK loop in each house are shown where (8) can be seen to be the common digit in r28, c28. Note that there is only one given for (8) at r6c6
SK Loop: 
(78=69)r79c2 - (69=58)r8c13 - (58=12)r8c79 - (12=48)r79c8 - 
(48-69)r13c8 - (69=58)r1c79 - (59=12)r2c13 - (12=78)r13c2 - Loop
=> 13 Eliminations: (58)r2c4, (58)r2c5, (48)r4c8, (78)r5c2, (48)r5c8, (58)r8c5, (5)r8c6
The loop is not an AIC. Two successive doublet terms will together hold both digits allowing them to be eliminated from the other cells in the same house. 
Proof of SK Loop Eliminations: Show In each of the nodes two digit doublets compete for two cells and for each pair of successive nodes one of these doublets will be repeated. 
Taking the first two nodes
 (78=69)r79c2 - (69=58)r8c13
if (78)r79c2 hold N truths then (69)r79c2 must contain (2-N) truths which will limit (69)r8c13 to N truths so forcing (58)r8c13 to hold at least (2-N) truths. 
As the loop is continuous, these restrictions will carry all the way round the loop so the final term on the second line, (78)r13c2, will be forced to hold at least (2-N) truths. But it can't hold any more than that because it links with the first term. Consequently all the left hand terms in the nodes will hold (N) truths and all the right hand ones will hold (N-1) truths. The possible outcomes are that all terms hold 1 truth or they alternately hold 2 and 0 truths. 
Regardless of which of these cases applies, two successive nodes must together contain both the digits in each doublet allowing them to be eliminated from the other cells in the same house.
Alternative Loop
(7=689)r79c2 - (689 =5)r8c13 - (5=128)r8c79 - (128=4)r79c8 - 
(4=689)r13c8 - (689=5)r2c79 - (5=128)r2c13 - (128=7)r13c2 - Loop
=> 15 Eliminations: (8)r1c3, (8)r1c7, (5)r2c4, (5)r2c5, (8)r3c1, (8)r3c9, (4)r4c8, (7)r5c2, (4)r5c8, (8)r7c1, (8)r7c9, (5)r8c5, (5)r8c6, (8)r9c3, (8)r9c7
Like the SK loop this is not a pure AIC as the triplet terms are considered true when they contain two truths and false when they contain one truth (the minimum they must hold). Two successive triplets will together hold all three digits allowing them to be eliminated from any other cells in the same house.  
The two loops give a mix of common and unique eliminations and it appears that the alternative loop is more productive. However they produce different follow-on eliminations after which the end results are identical. 
Follow-on eliminations after the SK Loop: (8)4Fish:c2458r1379 => r1c37,r3c19,r7c19,r9c37 <> 8    (8 eliminations)
Follow-on eliminations after the alternative loop: (8)4Fish: c2458r1379 => r2c45,r45c8,r5c2,r8c5 <> 8   (6 eliminations)
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- Reduced grid:
 *-----------------------*-----------------------*-----------------------*
 | <9>    278    457     | 2568   2358   3567    | 345    468    <1>     |
 | 258    <3>    158     | 1269   129    <4>     | 589    <7>    568     |
 | 457    178    <6>     | 1589   13589  13579   | <2>    489    345     |
 *-----------------------*-----------------------*-----------------------*
 | 4678   <5>    14789   | <3>    149    <2>     | 14789  169    4678    |
 | 23478  129    134789  | 1459   <6>    159     | 134789 129    23478   |
 | 2346   1269   1349    | 149    <7>    <8>     | 1349   <5>    2346    |
 *-----------------------*-----------------------*-----------------------*
 | 357    789    <2>     | 14589  134589 1359    | <6>    148    457     |
 | 568    <4>    589     | <7>    129    169     | 158    <3>    258     |
 | <1>    678    357     | 24568  23458  356     | 457    248    <9>     |
 *-----------------------*-----------------------*-----------------------*
(12=
69)r56c2 - (69)r79c2 = (69)r8c13 - (
69=12)r8c56 - (12)r8c79 = (12)r79c8 -
(12=
69)r45c8 - (69)r13c8 = (69)r2c79 - (
69=12)r2c45 - (12)r2c13 = (12)r13c2 - Loop 
This is a version of the SK loop with four key terms underlined. It uses only the four tell-tale digits that are givens in the pivot boxes. 
If the terms on the left contain two truths then (9) is forced true in r8c5 and r2c5 because (6) must occupy r8c6 and r2c4 
If the terms on the right contain two truths them (9) is forced true in r8c5 and r2c5 because (6) must occupy r6c2 and r4c8  
Because of these contradictions the each doublet is a STP (Single Truth Pair) and this will also apply to the usual version of the loop.  
Another consequence is that (1269)r2c45,r8c56 and (1269)r56c2,r45c8 must be PLQs (Pattern Locked Quads).
Together these derived inferences can simplify the rest of the solution considerably.  
The givens in boxes 2,4,6 & 8 follow a classic pattern as no two of them are in sight of each other. Now in box 5, the remaining non-loop box there is a given for one of the tell-tale digits (6), which is out of sight of all of these which signals that STPs will exist 
(if two adjacent boxes in the loop contain all four of the tell-tell digits as givens). STPs also frequently arise when the puzzle contains a JExocet too, but the proof is different.
Champagne before I make any further observations I would like know to if you have any examples of V-Loops that can't also be considered as SK loops? There are a number of ways of composing alternative loops but they all produce the same reductions as far as I can see. 
DPB
TAGdpbSKLoop
[Edit] Missing underlining in the third loop corrected, extra blue condition added for STPs to exist.