Rasputina creates ice walls on terrain that is uneven (say a hill). One of the markers is on the hill, the other is flat but they touch bases. \ _
Each have a height of 5 and are impassible and block LOS. A model of Ht 2 is behind the slanted marker. This model is not on the hill and is perpendicular to the table (not parallel to the slanted marker). No LOS can be drawn between the bases of the target (behind the marker) and the shooter. The top of the Ht 2 model is visible if the Ht of the slanted marker extends at the slant but not if it extends perpendicular to the table.
Does the height extend perpendicular to the base or to the horizontal of the ground?
Could you draw LOS to that model if it's base is behind the slanted marker and the height of the marker extends perpendicular to the base (so the height is also slanted)?
If the height is perpendicular to the table it would completely hide the model. If the height is slanted (perpendicular to the marker) then the top of the Ht 2 model is "visible" because the Ht 5 marker is extending at a slant.
Question
hofzinser
Here is the scenario;
Rasputina creates ice walls on terrain that is uneven (say a hill). One of the markers is on the hill, the other is flat but they touch bases. \ _
Each have a height of 5 and are impassible and block LOS. A model of Ht 2 is behind the slanted marker. This model is not on the hill and is perpendicular to the table (not parallel to the slanted marker). No LOS can be drawn between the bases of the target (behind the marker) and the shooter. The top of the Ht 2 model is visible if the Ht of the slanted marker extends at the slant but not if it extends perpendicular to the table.
Does the height extend perpendicular to the base or to the horizontal of the ground?
Could you draw LOS to that model if it's base is behind the slanted marker and the height of the marker extends perpendicular to the base (so the height is also slanted)?
If the height is perpendicular to the table it would completely hide the model. If the height is slanted (perpendicular to the marker) then the top of the Ht 2 model is "visible" because the Ht 5 marker is extending at a slant.
Edited by hofzinserClarified question
Link to comment
Share on other sites
6 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.