

Sorry is this had been treated somewhere:
About the LOS of the foot artillery, there is only one case confusing: A range 5 special case that we use to declared BLOCKED (no LOS) ... Are we right? Here it is:
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Guido Gloor
Switzerland Ostermundigen Bern
The statement below is false.
The statement above is correct.

Rules page 9:
rules wrote: If the imaginary line runs along the edge of one or more hexes that contain obstructions, line of sight is not blocked unless the obstructions are on both sides of the line. The case you display is even less severe than the LOS line going along an edge of a hex, you're merely touching the corner. I see no argument for LOS being blocked in this case.



Well ... with respect: are you sure?
Because I saw some discussions about this mathematic case. Any other opinions (even on the same side)? GG

Aron
Netherlands
Life is game, you gotta enjoy the journey.
Life may knock you down sometimes. Its up to you if you stand up and live your life to the fullest.

GGleize wrote: Well ... with respect: are you sure? Because I saw some discussions about this mathematic case. Any other opinions (even on the same side)? GG
Nothing sure. But my opinion is that A can see B in this situation. Anybody have arguments against this? (not that I have arguments for, its just a feeling).



Two points that made us say no LOS:
1) "If the imaginary line runs along the edge of one or more hexes that contain obstructions, line of sight is not blocked unless the obstructions are on both sides of the line" ... OK but our question is: Does that corner of hex CUT OR NOT THE LINE? We used to say YES but we might be wrong: that was my initial question to you all! It's not a point of seing or not the other hex because sometime the blocking hexe let see a little part ... but the wargame hex board tradition need a free line between the two CENTERS of the hexes.
2) The second reason we would be sorry if you all say that the LOS is OK here is that: To avoid long discutions we used a simple rule. When it's not in a straight line, both hexes must be one the SAME SIDE of a possible blocking hex to have a LOS (see under)!
It works perfectly at ranges of 2,3 and 4. It works at range 5 but has only one unclear case (the case we speak about). And it sure will not work at range of 6 or more (but we don't care because no such range in CCN). So I was dreaming of you all to agree that the only unclear case of the CCN range foot artillery of 5 max would be a NO LOS to make our easy method be 100% efficient! Please don't tell me the opposite ...!

Guido Gloor
Switzerland Ostermundigen Bern
The statement below is false.
The statement above is correct.

Yes, I am absolutely sure. First, it's important to realize that you can divide a hex into six triangles, each of which has three equallength sides. Then, the problem becomes trivial:
As you see, there are essentially three segments of this line, each of them crosses four of those triangles and then arrives at a junction between them again. In these junctions (marked with red circles), the line shares exactly one point with the triangle grid, it goes exactly through the crossing of the lines.
The point you are interested in happens to be exactly one of those junctions. So the line, in that spot, shares exactly one geometrical point with the hex border. And that's why I quoted the rule that I quoted: If the line only shares points with the hex border (as it does when it goes along the side of the hex, plenty of them actually) and does not go through the area of the hex at any point, there is LOS.
Your second reason unfortunately is a rule of thumb only and might work for other cases, but obviously produces a geometrically wrong interpretation in this case.

brian
United States Cedar Lake Indiana

If you can't figure it out visually, use math. If LOS runs through a single point or along the edge of a hex and only has blocking LOS on one side, then you DO HAVE LOS.
Hexes can be converted to a graph. You can determine where the point in question is at by taking an arbitrary height of 1 across the hexagon making the apothem 1/2 (the inradius or radius of the inside circle).
You can then take the two points we have of the line and calculate out the slope of the line. If you know the x, y coordinates of the point in question, you can check your y value to calculate the x value and vice versa.
If both methods give you the same result, then the line and the vertex (the point in question) are the same point and LOS is not blocked.
I had to brush off my math memory and calculate on a few scrape of paper, but I can say they share the same point. The line does NOT intersect the area of the hexagon and LOS is present in your third (right most) example.

brian
United States Cedar Lake Indiana

Or I guess a more detailed graphic would have been easier....
At least the math supports the conclusion too!

Read the rulebook, plan for all contingencies, and…read the rulebook again.
United States Austin Texas

haslo wrote: Yes, I am absolutely sure. First, it's important to realize that you can divide a hex into six triangles, each of which has three equallength sides. Then, the problem becomes trivial: As you see, there are essentially three segments of this line, each of them crosses four of those triangles and then arrives at a junction between them again. In these junctions (marked with red circles), the line shares exactly one point with the triangle grid, it goes exactly through the crossing of the lines. The point you are interested in happens to be exactly one of those junctions. So the line, in that spot, shares exactly one geometrical point with the hex border. And that's why I quoted the rule that I quoted: If the line only shares points with the hex border (as it does when it goes along the side of the hex, plenty of them actually) and does not go through the area of the hex at any point, there is LOS.
That's a helpful diagram, haslo! I agree that LOS is clear in the original situation.
Since you are allowed to draw LOS to a target down a hex spine so long as as there is no blocking terrain on both sides of the line along its course, what would you say to LOS being blocked if both hexes I4 and J2 had blocking terrain?

Guido Gloor
Switzerland Ostermundigen Bern
The statement below is false.
The statement above is correct.

BradyLS wrote: Since you are allowed to draw LOS to a target down a hex spine so long as as there is no blocking terrain on both sides of the line along its course, what would you say to LOS being blocked if both hexes I4 and J2 had blocking terrain? I agree, LOS would have to be blocked then



To Guido & Brian: very clear demonstration.
LOS is open. End of my question. TY very much Generals!
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