I don‘t think that 4♣ unambiguously sets ♣ as trumps, after all 2♣ doesn‘t deny ♠ support. In order for 4♣ to be effective it should draw a very clear and narrowly defined picture. The partneship should define what the picture looks like. You can define other picture bids like 4♦/♥ and bid 3♣ with an album of all the other support pictures.

Call it what you like: Lebensohl or GB2N.

Nice to read D-I without me having to write it.

Same for me

We play Michaels as split range, omitting 5-6 loser hands. If I choose 2♦, I‘d have to bid 4♦ showing 6♠-5♥ and upper range. I‘d prefer 1♠ however, the hand is borderline, but ♠s are much better. Then 1N-3♥-3N-4♠.
Edited: 1N (10-14, no fit), not 2N (inv+, defensive 4crd raise) of course.

XX to show 1st round control is not unusual. 4♥ would then show ♥-control plus help in ♦ (Q or stiff) and pass for xx(x).

No problem for me: 3♣ shows 5-5.

Correct.

Not just strong, FG according to OP.

Definetly 1♣. If 2♣ is FG, I guess you played some game, maybe you were lucky, but there is no adjustment for luck. I‘m not even sure if there should be an adjustment if you stop below game.

4 bids? 4♥-P-P-5♥;P-6♠-P-7♠!

I‘m sure I will be corrected if I‘m wrong about Acol.

Of course it was natural in 1947. Even today I would expect it to be natural (and forcing) in Acol-land (my vote). In Germany (and probably elsewhere) it shows ♥-stop but no ♠-stop IMO.

Corrected!

I voted other because the main problem is the enormous range of any raise in a SNT-based system. Assuming that most people open a 4333 12 count, the minimum is 11+ Kleinman points (deduct 2/3 for the 4333). Consequently the maximum for a single rsise can hardly be more then 16-. A jump raise then starts at 16, but the maximum must be a 19- if not more.
W has 14hcp, deduct 1 for the stiff J and 1/3 for the unsupportet J, giving 13- Kleinman points. According to Kaplan we now add 3 fit points for the stiff and come to 16- „my points“. A maximum single raise. As for E, I count 6hcp substract 1/3 for the unsupported J and add 1/3 for the A and the Supported T, giving 6+ Kleinmann points. No adjustment for a 5crd because of lack of quality. As long as I can be sure that pard has 4crd support I can add 1 point for a doubleton in a 9crd fit or a total of 7+ „my points“. This is an invite opposite a range of 16-19-, but unfortunately there is no way to invite anymore. Note that giving W ♦Q in stead of J or even ♥A in stead of K hardly improves the chances of making 4♠ even though W then has a clear jump raise given the methods used. Imagine playing WNT. Now a single raise has a range of 14+-17+ and a jump raise 18- -19+ and E has no difficulty bidding game with his 7+. In another post I wrote „the real strength of WNT is when you DON‘T open 1N“. Here is an example why.
Edited: correction because of W hcp.

Kleinman points: http://www.jeff-goldsmith.org/bridge/dk.doc

Craig, I hit the „add comment“ button before I was ready and it seems you replied after reading only the first sentence. Not your fault of course.

“that is how the NLTC method works.” is not an explanation that will convince others to adopt it. Worse still, if you don‘t know why you will not be able to make the necessary adjustments because of extra information. Your „The “problem” with NLTC here is (partially) that the power stiff ♥ opposite 0 wastage in the other hand is not properly reflected.“ is what I tried to explain above. I‘m not a fan of NLTC or Klinger LTC. A couple of years ago I analysed some 4k hands played in our club and compared the correlation between calculated and DD # of tricks for NLTC, Goren point count and „my points“, where „my points“ got the best correlation. „My points“ is Danny Kleinman‘s plus adjustments for 5+ suits depending upon suit quality and fit points as described in „how to play winning bridge“, the first bridge book I ever read.But whatever you useif your scores will improve if you adjust according to information from the bidding, especially with respect to wasted values.

Compensating duplication of values. Example: Pard has Qxx counting as 2.5 losers. Opposite a stiff Qxx is the same as xxx or 3 losers. The opposite is counting Axx as 1.5, not 2 losers because the A improves pards Qxx.