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My third instruction has 6 pairs of letters and a full word after the first pair.

Fairfield - I only had 9 words but found I had missed the word at the end of the second instruction....

Got there in the end and enjoyed the challenge. Ginge thanks for explaining the "house" in 9ac.
Fairfield - with only 9 words, you may have fallen in to the same trap as me in the clue for 1d. I assumed the penultimate word was the one to be removed giving AY and making a word with the two letters of 29ac. I later realised it is the first word of 1d that is to be removed and used in full. The clue I'm not so sure about but the 3rd instruction reads better.

1d = CAVA (wine) + TIN (can) + A (lead to abysmal, ie first letter)

Am I being even more stupid than usual? I have arranged the completed grid so that the building and city are legible reading downwards in 3 columns, but the rest of the grid is meaningless. How does one "use all cells". I can't see where the demonstration is explained. I can't even see the point of the 3rd instruction. I'm obviously missing something vital - but what?

OK I now see the point of the 3rd instruction and have a grid that looks like the building. But is that all?

Escuan, there are so many weaknesses in this puzzle, both conceptually and practically; reading, if you have the patience, all the comments on this thread will give you a flavour.

I don't send in my entries, but did de-construct the grid to make something with both the building and city appearing in just one vertical line (apart from 'The', which Dryden earlier questioned).

The consensus is that L and S can both be on the ground (which demonstrates nothing, because the observer doesn't know at what time each object came to be on the ground; or that L & S can be in mid-air, equal distances below their dropping points. There are various obstacles here too, in that one of the objects seems then bound to obscure existing contents of a cell.

All the cells are used as part of the building, the name of the building, the place of the building, and objects which were probably never thrown off the building. And that's all there is...

Thanks for that masterful summary Meursault. My final problem has been that L and S started at different levels, so the experiment was flawed from the outset - L would arrive first as it didn't have so far to fall. I pity John Green having to cope with the various efforts at solution!

The more I think about it, there is in fact only one correct solution. If the objects are dropped from different heights the experiment becomes more difficult to measure, though not impossible, and it is crucial that there is accurate timing. To show both objects on the ground is therefore wrong : all that does is demonstrate gravity. The correct solution should show both objects in a corresponding intermediate state.

I am completely baffled by this. If I put the various sections (and where does it tell you to cut out these sectons?) on top of one another and 'slide' them so that the building and place read vertically in a straight line, then the representation of the building is wrong and the 'objects' start at different levels. However, if the sections are simply put on top of one another and not slid, but the whole construction is tilted, then the building is better represented and the objects would be on the same level. Or am I barking up the wrong tree completely?