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Posted on 2007-04-01
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Hi,

I've asked this question a few times but still can't understand the answers. I need a really dumbed down answer.

I have two points in 3D space:

Point1: (x,y,z)
Point2: (x,y,z)

I want to draw an arrow from point1 to point2. At point2 I'd like to draw a pyramid cap to mimic an arrow head. So I need to know how to generate an additional 4 points that are equally distributed around point 2 - if I connect them all to point 2 it will look like a pyramid whose tip is point2.

Can someone please tell me how to fill this in then (the four generated base points of the pyramid arrow head):

base_pt_0 = (x,y,z);
base_pt_1 = (x,y,z);
base_pt_2 = (x,y,z);
base_pt_3 = (x,y,z);

If it is any easier - I could also draw a cone shape for the arrow head - i guess just like 16 points or so evenely distributed around point 2.

Thanks
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Question by:DJ_AM_Juicebox
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Expert Comment

ID: 18834316
Did you not get the answer from the previous question?
It looks like you also want to move the points around point2 a little bit back toward point1, so add (point2-point1)*(1-(arrowhead length)/(line length)) to each of those points.
to add intermediate points, you can use
(Point1-Point2) × (Point0-Point2)  * sin(45º)
+
(Point1-Point2) × (PointA-Point2) * cos(45º)
for every 22.5º
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Author Comment

ID: 18834362
What is point0 again?
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LVL 84

Expert Comment

ID: 18834578
point0 can be any point not along the line through point1 and point2
perpendicular0 = (Point1-Point2) × (Point0-Point2)
perpendicular2 = perpendicular0 × (Point1-Point2) / |Point1-Point2|
pontx = point2+(point1-point2)*(length of cone)/|Point1-Point2| + sin(x)*perpendicular1+cos(x)*perpendicular2
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Author Comment

ID: 18836296
Ok so I get these 3 perpendiculars -

then you show me that I will get this 'pointx' -

do you mean that pointx is just any point that can be generated that will be at the base of the head (I just substitute in for 'x' a number between 0 - 360, depending on how many points I want in the base)?

Thanks
0

LVL 84

Expert Comment

ID: 18839930
yes
0

Author Comment

ID: 18861727
Hi ozo,

I finally got a chance to give it a try. A few questions:

1)   perpendicular1 = perpendicular0*(radius of cone)/|perpendicular0|
what do I do here when a component of perpendicular0 is zero?

2) perpendicular2 = perpendicular0 × (Point1-Point2) / |Point1-Point2|
here if I put parenthesis in the equation it should be:

perpendicular2 = (perpendicular0 × (Point1-Point2)) / |Point1-Point2|

I don't know if cross product has a higher precedence than division.

Thanks
0

LVL 84

Expert Comment

ID: 18862409
|perpendicular0| will be 0 only if Point0, Point1, and Point2 are all on the same line
which is why Point0 can't be on that line.
individual components of perpendicular0 can be 0, in which case the corresponding component of perpendicular1 will be 0

It doesn't matter where you put the parenthesis, the result will be the same.
0

Author Comment

ID: 18863161
ok dumb question, but is there some guaranteed way of picking a point not coliear to point 1 and 2?
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LVL 84

Expert Comment

ID: 18863185
At least one of Point1 + (1,0,0) and Point1+(0,1,0)
will not be colinear to point 1 and 2.
0

LVL 84

Expert Comment

ID: 18863418
If you take any three non-colinear points, at least one will not be not colinear to point 1 and 2
0

Author Comment

ID: 18864782
Hi ozo,

I just picked two points in a real stuation for my arrow tail and head, they are:

Point1(0, 159, 83);
Point2(0, -0.3, 83);

Now I'm just manually choosing a non collinear point for poin0 of:

Point0(10, 10, 0);

but still it gives me zero in the components with this formula:

perpendicular0 = (Point1-Point2) × (Point0-Point2)

perpendicular0 will turn out to be:

perpendicular0 = (13320, 0, 1600)

am I doing something wrong?

Thanks
0

LVL 84

Accepted Solution

ozo earned 500 total points
ID: 18868112
I get (13221.9, 0, 1593)
but that's close enough
A zero component is not a problem as long as they are not all zero
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