Plotting Arrows

I need to be pointed in the right direction. In a program that I'm writing I need to draw an arrow. I have a line that can be any random angle. What formula can I use to draw two other lines that begin at an X2,Y2 point and are slanted at a 45 degree angle for instance. The endpoints of the lines that make up the arrow does not matter.

For example:

X1=0
Y1=0
X2=10
Y2=10

How can I plot an arrow off of theese points using a formula?

Thanks for any help.

Well, I figured out in a round about way to do it. It's not a formula like I was looking for but for those who are interested, here it is.

VB Code:

x1=0:y1=0:x2=300:y2=400
 
    xLineDelta = X2 - X1
    yLineDelta = Y2 - Y1
 
    xLineUnitDelta = xLineDelta / Sqr((xLineDelta * xLineDelta) + (yLineDelta * yLineDelta))
    yLineUnitDelta = yLineDelta / Sqr((xLineDelta * xLineDelta) + (yLineDelta * yLineDelta))
 
    '(xBase,yBase) is where arrow line is perpendicular to base of triangle.
    HeadLength = 15 'The Length Of The Arrow Head
    xBase = X2 - Round(HeadLength * xLineUnitDelta)
    yBase = Y2 - Round(HeadLength * yLineUnitDelta)
 
    xNormalDelta = yLineDelta
    yNormalDelta = -xLineDelta
    xNormalUnitDelta = xNormalDelta / Sqr((xNormalDelta * xNormalDelta) + (yNormalDelta * yNormalDelta))
    yNormalUnitDelta = yNormalDelta / Sqr((xNormalDelta * xNormalDelta) + (yNormalDelta * yNormalDelta))
    
 
    '// Draw the arrow tip
    Me.Line (X1, Y1)-(X2, Y2)
 
    'here is one side of the arrow
    'x1=xBase+Round(HeadLength * xNormalUnitDelta)
    'y1=yBase + Round(HeadLength * yNormalUnitDelta)
    Me.Line (xBase + Round(HeadLength * xNormalUnitDelta), yBase + Round(HeadLength * yNormalUnitDelta))-(X2, Y2)
 
    'here is the other side of the arrow
    'x1=xBase+Round(HeadLength * xNormalUnitDelta)
    'y1=yBase + Round(HeadLength * yNormalUnitDelta)
    Me.Line (xBase - Round(HeadLength * xNormalUnitDelta), yBase - Round(HeadLength * yNormalUnitDelta))-(X2, Y2)

Thanks

OK, when drawing lines it is always a good idea to look at the angle formed between the line and the x-axis.
this way, if you want a line at T degrees, length L from the x-axis starting at X1,Y1, the line is:
(X1,Y1)->(X1+R*cos(T), Y1+R*sin(T))

Then, as you are now wanting two lines 45 degrees off the endpoint, the new lines are at 45+T and 135+T degrees from the X-Axis (or pi/4 +T and 3Pi/4 +T radians)

You can use the same formula then to draw the new lines, with starting point X2=X1+R*cos(T) and Y2=Y1+R*sin(T)

Finally, i'm not too sure about what happens with angles with the same tan value (like, 30 and 210 degrees) whether this works fine, or whether some if-statements must be added to see if the angle is < or >=180 degrees.

I think this should be correct in all instances.

Referring to the drawing, (your arrow is blue):

http://www.vbforums.com/attachment.p...postid=1473475

alpha is 45 deg. in your case, but I have written the general case.

theta is the angle between the (x₁,y₁) to (x₂,y₂) segment and the x axis. In the drawing it is > 180 deg. because the arrow is pointing downwards and to the left.

The coordinates you want to figure out are (x₃,y₃) and (x₄,y₄).

Then:

x₃ = x₂ + cos(Beta2)
y₃ = y₂ + sin(Beta2)
x₄ = x₂ + cos(Beta1)
y₄ = y₂ + sin(Beta1)

and cos & sin of Beta2 and Beta1 as a function of alpha and theta are calculated by means of elemental trig. formulas:

Beta2 = theta - (Pi - alpha)
Beta1 = theta + (Pi - alpha)

cos(Beta2) = sin(theta)sin(alpha) - cos(theta)cos(alpha)
sin(Beta2) = -sin(theta)cos(alpha) - cos(theta)sin(alpha)
cos(Beta1) = -sin(theta)sin(alpha) - cos(theta)cos(alpha)
sin (Beta1) = -sin(theta)cos(alpha) + cos(theta)sin(alpha)

I have written a demo project where I have made a few changes to take user scales into account, otherwise if the vertical and horizontal scales are not the same your arrow could come out quite out of this world.

Here's a new drawing I've used in relation to the code.

http://www.vbforums.com/attachment.p...postid=1500193

Couldn't you have an X and Y range?

If you did then you could find the position by (this is for X):
(X - Xmin) / (Xmax - Xmin) = decimal percent. of placement inside pic
where Xmin <= X <= Xmax
then
Pic1.left + round( dec. percent * Pic1.Width) = X coordinate

This way, your graph is scalable. :)

NOTE: Please check to make sure I'm right. Also, sorry if I am reiterating anything here. I did not get a chance to look at the attachment.

Quote:

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Originally posted by Darkwraith
Couldn't you have an X and Y range?

If you did then you could find the position by (this is for X):
(X - Xmin) / (Xmax - Xmin) = decimal percent. of placement inside pic
where Xmin <= X <= Xmax
then
Pic1.left + round( dec. percent * Pic1.Width) = X coordinate

This way, your graph is scalable. :)

NOTE: Please check to make sure I'm right. Also, sorry if I am reiterating anything here. I did not get a chance to look at the attachment.

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At first glance I'd say this is more or less what I do. See the attachment and take a look at the code.