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 Post subject: Attack Vector Tactical pg 22 Segment 4
PostPosted: Mon Jan 29, 2018 5:11 pm 
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Joined: Tue Mar 07, 2017 7:08 pm
Posts: 7
Location: Houston, Texas
Quick question. On page 22 of the main Attack Vector Book, in segment 4, it says to apply thrust, to "check off another thrust". I don't know what to mark here. I attached a jpeg of the text I am reading below.

Just before that, On page 21, a little at the end of A5.342 (segment 3) there is this information:

"Some of the acceleration dots will accumulate in the horizontal direction...and some in the + direction"

Now we go to resolve thrust in A5.343

At the end of this segment, there is now a dot in C.

In Segment 4, we gain an acceleration dot in C and an acceleration in +.

Quick question: Why did we not have a dot in "+" also in A5.343?

Just after that, in section A5.351,I'm not sure what to mark in "so after marking the thrust chart". There are two green boxes now, not a green and then a white. What changes when I go from a white & green box to two green boxes?

I uploaded a photo so you can see where I am at on my SCC.


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 Post subject: Re: Attack Vector Tactical pg 22 Segment 4
PostPosted: Mon Jan 29, 2018 7:15 pm 
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Joined: Tue Mar 07, 2017 7:08 pm
Posts: 7
Location: Houston, Texas
Hi all,

I'm still having some trouble on page 22 of the main Attack Vector Manual. This questions are about A5.365.

It says “This leaves us a vector with a 1+ to resolve.”
8 goes into 1 zero times with a remainder of 1.

I don’t see where the “8” came from. Is it from the fuel? (A5.362 we have burned 8 units of fuel)

Why do we do 1 divided by 8? I understand that the remainder will be 1 from this math operation.
I understand the combined horizontal components add up to 6.

Treat this as the first part of a remainder split and the 1 as the second part, and look up that split....

The pink column suggests that the split to use was the 6|1 split for the Vertical Remainder Grid. At the very bottom of this paragraph though, it says “your combined vector is a 4|2|1 split.

The 4|2|1 split is shown in the top row of the Movement Grid figure. But we filled in the pink column with a 6|1 split. So I’m confused about that.


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 Post subject: Re: Attack Vector Tactical pg 22 Segment 4
PostPosted: Tue Mar 06, 2018 5:58 pm 
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Joined: Fri Jul 10, 2009 12:03 pm
Posts: 125
Location: Austin, TX
Did you ever resolve this? You seem to be conflating several separate concepts here, so hopefully we can set you straight. The thrust chart is extremely information-dense, so this confusion seems pretty reasonable.

The thrust chart for a given level of thrust is made of 8 boxes. Each box is divided into 3 colored zones, each of which may contain displacement arrows; plus the fuel consumption in the middle. The left-hand colored zone (white/yellow/orange) describes the thrust accumulated in the current direction, for when you're thrusting in the amber or purple rings, parallel or perpendicular to the plane of the map. The right-hand colored zones (white/lime/green) describe the thrust accumulated in the primary (top) and secondary (bottom) directions, for when you're thrusting in the blue or green rings, at an angle to the plane of the map. The dots in the middle describe the amount of fuel (propellant, technically) consumed in this segment of thrust.

In this example, you are thrusting at Thrust 4 in the C+ direction. This is in the blue ring, so you use the right-hand divided part of the thrust chart. Your primary direction (where you're putting "most" of the thrust) is C, and the secondary direction (where you have "some" of the thrust) is +. (If you were thrusting in the green C++ direction, these would be reversed.) The first segment of the thrust chart has lime in the top half and white in the bottom half, which means you get one thrust dot in the primary direction (C) and nothing in the secondary direction (+) this segment. You also spend one dot of fuel. Your photo has this right.

For the second segment, you mark off the next box (or double-half-box) in the thrust chart. This box has lime in both halves, so you get a thrust dot in both C and +, leaving you with two dots in C and one in +; and you spend one more dot of fuel.

If you left the torch on without pivoting for 8 segments, you would continue this way, eventually accumulating 7 dots in C and 4 in +, while spending 8 fuel dots. If you check the RALT, you see that 7 out and 4 up would be a range of 8, so you've changed your total velocity by 8 hexes per turn, as expected from the amount of fuel it cost you. Thrust 4 represents an acceleration of 8 hexes per turn per turn.

However, we instead decide to shut down the torch after only 2 segments. This triggers a thrust break, which means you have to take whatever dots you've accumulated and merge them into your actual vectors, then redo your movement grid (the subject of your second question). I'm assuming that you managed add the dots to the vectors and simplify your vectors correctly, to end up with a final vector of 4B/2A/1+ per turn. Now we have to fill in the movement grid to represent this vector.

First we look at the horizontal part, 4B/2A per turn. Divide each of these by 8 (the number of segments in a turn). The quotients (0B/0A) go down the left part of the yellow and brown columns of the movement grid, and we consider only the remainders: 4B/2A. (This step didn't do anything in this case because we're not going very fast.) Now look at the horizontal movement grids, and pick the one labeled 4/2. Fill in the right side of the yellow and brown columns of the movement grid according to the marks on the reference card; so you get an extra B in segments 2, 3, 6, and 7; and an extra A in segments 4 and 8; added to the 0 hexes you move every segment (from the "divide by 8" step earlier). If you add all of these up, you get 4 total hexes per turn in B and 2 hexes per turn in A, as expected. (This whole process is more complicated if you're moving faster than 8 hexes per turn in some direction.)

Next we move on to the vertical movement grids. The first step is to add together your horizontal movement directions to get the total horizontal movement: 6. So your total horizontal/vertical vector is 6H/1+ per turn. Again, you divide by 8 segments/turn and put the quotient (0) down the left half of the pink column on the movement grid. Then you take to the 6/1 grid on the vertical movement reference, and copy the right half of that onto the right half of the pink column. (The left half should match the combination of the extras on the yellow and brown columns, but it doesn't matter at this step.) You therefore get an extra + movement in segment 5 (I think, I don't have the reference with me; it might be segment 1), for a total of 1 + movement per turn. In the end your movement grid should look something like this:
Code:
  +---------+---------+---------+
  |  4  | B |  2  | A |  1  | + |
  +---------+---------+---------+
  |  0  | 4 |  0  | 2 |  0  | 1 |
1 |     |   |     |   |     |   |
2 |     | 1 |     |   |     |   |
3 |     | 1 |     |   |     |   |
4 |     |   |     | 1 |     |   |
5 |     |   |     |   |     | 1 |
6 |     | 1 |     |   |     |   |
7 |     | 1 |     |   |     |   |
8 |     |   |     | 1 |     |   |
  +---------+---------+---------+

If you need a little practice, try the movement grid with a much more extreme vector, like 13F/6E/11-. If you get to the end with the right total in each colored column of the movement grid, you probably did it right.


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 Post subject: Re: Attack Vector Tactical pg 22 Segment 4
PostPosted: Thu Apr 12, 2018 8:52 pm 
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Joined: Wed Sep 07, 2005 11:28 am
Posts: 1329
Location: El Segundo, CA
Nathan Bezayiff wrote:
Hi all,

I'm still having some trouble on page 22 of the main Attack Vector Manual. This questions are about A5.365.

It says “This leaves us a vector with a 1+ to resolve.”
8 goes into 1 zero times with a remainder of 1.

I don’t see where the “8” came from. Is it from the fuel? (A5.362 we have burned 8 units of fuel)

Why do we do 1 divided by 8?


There are 8 segments in each turn. If you have a vector of 8 in one direction, you're moving one hex per segment in that direction. If your vector is 9, you'll move 1 hex per segment, plus 1 additional hex on one of those segments. That's obvious, but the formula is "divide by 8, looking for a remainder."

If your vector is 21, you divide 21 by 8 to get 2r5 (2 with a remainder of 5). You'll move 2 on every segment, and an additional 1 on five of the segments. The movement grid with tell you which five segments you'll move those additional hexes.

(Paul did a good job answering. This is just a small additional clarification.)

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