Im trying to do a graph from evalued math function and this is last think I need to do. I have graph with limit coordinates -250:-250 left down and 250:250 right up. I have Y-limit function, which is defined as -10:10, but it could be user redefined and if it is redefined, I need to calculate new coordinates.
I have now field of y-coordinates with 20000 values and each of is multiply by:
ratioY = 25 / (fabs( up-limit - down-limit ) / 20) which will make coordinates adapt for new Y-limit (if limit is -5:5, graph looks 2x bigger), this works good, but now isnt graph exactly where it should be (see pictures). Simply 25 is multiplied for postscript coordinates and (up-limit - down-limit) / 20 is ratio for "zooming" Y coordinates. This works fine.
Now Im trying to "move coordinates" which will subtracted from revaluated value:
ycoor = (ycoor * ratioY) - move-coorY ;.
Now I have something like this:
move-coorY = 25* ( ( up-limit - down-limit) /2 );
and it doesnt work correctly. I need to do sin(0) start from 0.
This is a correct graph which is -10:10
(source: matematika.cz)
This is a bad graph which is -5:10
(source: matematika.cz)
Maybe its easier not to do this with fixed numbers (like your ratioY) but with two different coordinate systems. Physical coordinates are in your problem domain, i.e. they are the real values of your sine curves. Logical coordinates refer to the device, in your case they are the point values in Postscript, but they might be pixels on a HTML canvas or whatever.
I'll denote the physical coordinates of the first axis with a small x and the corresponding logical coordinate with a capital X. In each coordinate system we have:
Lower bound: x_min, X_max
Upper bound: x_max, X_max
Range: dx = x_max - x_min
dX = X_max - X_min
Then you can calculate your logical coordinates from the physical ones:
X(x) = X_min + (x - x_min) * dX / dx
This also works vice versa, which is not an issue for Postscript files, but max be useful for an intractive canvas where a mouse click should yield the physical coordinates.
x(X) = x_min + (X - X_min) * dx / dX
In your case, the ratio or scale factor is dX / dx, which you can calculate once for each axis. Let's plot the first point with y == 0 in your first graph:
y_min = -10
y_max = 10
dy = 20
Y_min = -250
Y_max = 250
dX = 500
Y(0) = -250 + (0 - (-10)) * 500 / 20
= -250 + 10 * 500 / 20
= 0
In the second graph, the logical coordinates are the same, but:
y_min = -5
y_max = 10
dy = 15
Y(0) = -250 + (0 - (-5)) * 500 / 15
= -250 + 5 * 500 / 15
= -83.3333
If you change the range of your graph, e.g. from (-10, 10) to (-5, 10), just adjust the physical coordinates. If you resize your graph, change the logical coordinates. (Also, calculating the point in the graph is the same as calculating the position of the tick mark for the axis. Strangely, you got the tick marks right, but not your graph. I think your problem was to account for the non-zero lower bound in both graph and curve data.)
Also, it's probably better to re-evaluate the logical coordinates when printing instead of transfroming them from a previous plot. You can do that on the fly, so that you only need to keep the physical data in an array.
(And lastly, I'll admit that I'm not entirely sure these two kinds of cooirdinates are called physical and logical. I know these terms are used, but it may be the other way round or they might even mean sonething different altogether.)
My friend did a well yob for me and programmed this...
double zeroPosition(double startY, double endY){
double range = endY - startY;
double topSize = endY / range;
return 250.0 - 500 * topSize;
}
This will calculate position of zero, which I just add to my Y position with ratio and It works exactly how I need!
But thanks M Oehm ;)
Related
I want to plot an NxN array of circles. Just to visualize, I attached an image of what I want to achieve. I'm new in MatlLab so I tried to plot a single circle first, and here is the sample code below:
n = 2^10; % size of mask
M = zeros(n);
I = 1:n;
x = I - n/2; % mask x - coordinates
y = n/2 - I; % mask y - coordinates
[X,Y] = meshgrid(x,y); % create 2-D mask grid
R = 200; % aperture radius
A = (X.^2 + Y.^2 <= R^2); % Circular aperture of radius R
M(A) = 1; % set mask elements inside aperture to 1
imagesc(M) % plot mask
axis image
I really don't have any idea on how to plot a 2D-array of circles. The distance between two circles is two radii. I need this for my research. Hoping anyone can help.A 4 x 4 array of circles.
If you just want to plot a set of circles, you can use the rectangle function within a loop
If in the call to rectangleyou set the Curvature property to 1 it will be drawn as circle (ref. to the documentation).
In the loop you hav to properly set the position of each rectangle (circle) by defining its lower left coordinates along with its width and height.
Having defined with n the number of circles and with d their diameter, you can compute the set of lower left coordinates as:
px=linspace(0,d*(nc-1),nc)
py=linspace(0,d*(nr-1),nr)
The black background can be either obtained by setting the color of the axes or by plotting another rectangle.
Then you can set the xlim and ylim to fit with the external rectangle.
You can also make the axex invisible, by setting its â—‹Visibleproperty tooff`.
Edit #1
Code updated to allow drawing a user-defined number of circles (set the number of rows and the number of columns)
% Define the number of circles
% Number of rows
nr=8
% NUmber of column
nc=8
% Define the diameter of the circle
d=6
% Create an axex and set hold to on
ax=axes
hold on
% Evalaute the lower left position of each circle
px=linspace(0,d*(nc-1),nc)
py=linspace(0,d*(nr-1),nr)
% Plot the background rectangle
rectangle('Position',[px(1),py(1),d*nc,d*nr],'Curvature',[0 0],'facecolor','k')
% Plot all the circles
for i=1:length(px)
for j=1:length(py)
rectangle('Position',[px(i) py(j) d d],'Curvature',1,'facecolor','w')
end
end
% Set the aspect ratio of the axes
daspect([1 1 1])
% Set the XLim and YLim
xlim([px(1) d*nc])
ylim([py(1) d*nr])
% Make the axes invisible
ax.Visible='off'
Edit #2
Aternative solution to address the OP comment:
If I want to make the axes fixed, say I want a meshgrid of 1024 by 1024, (the image size is independent of the circle radius) how do I incorporate it in the code?
If you want to use a fixed (1024 x 1024) mask on which to plot the circles, starting from the cde you've posted in the question, you can simply do the following:
define the number of circles you want to plot, on a (1024 x 1024) mask they can be 2, 4, 8, ...
define a basic mask holding just one circle
identify the points inside that circle and set them to 1
Replicate (using the function repmat the basic mask accoding to the numner of circles to be plotted
The implemntation, based on your code could be:
% Define the number of circles to be plotted
% on a (1024 x 1024) mask they can be 2, 4, 8, ...
n_circles=4
% Define the size of the basic mask
n0 = (2^10)/n_circles;
% Initialize the basic mask
M0 = zeros(n0);
% Define the x and y coordinates of the basic mask
I = 1:n0;
x = I - n0/2;
y = n0/2 - I;
% Create the mask
[X,Y] = meshgrid(x,y);
% Define the radius of the basic circle
R = n0/2;
% Get the indices of the points insiede the basic circle
A = (X.^2 + Y.^2 <= R^2);
% Set basic mask
M0(A) = 1;
% Open a FIgure
figure
% Replicate the basic mask accoding to the numner of circles to be plotted
M=repmat(M0,n_circles,n_circles);
% Display the mask
imagesc(M)
% Set the axes aspect ratio
daspect([1 1 1])
Here is the graph I currently have
:
The Dotted Blue line represented the y value that corresponds to the x value I am looking for. I am trying to find the x values of the line's intersections with the blue curve(Upper).Since the interesections do not fall on a point that has already been defined, we need to interpolate a point that falls onto the Upper plot.
Here is the information I have:
LineValue - The y value of the intersection and the value of the dotted line( y = LineValue)
Frequency - an array containing the x value coordinates seen on this plot. The interpolated values of Frequency that corresponds to LineValue are what we are looking for
Upper/Lower - arrays containing the y value info for this graph
This solution is an improvement on Amro's answer. Instead of using fzero you can simply calculate the intersection of the line by looking for transition in the first-difference of the series created by a logical comparison to LineValue. So, using Amro's sample data:
>> x = linspace(-100,100,100);
>> y = 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
>> LineValue = 0.8;
Find the starting indices of those segments of consecutive points that exceed LineValue:
>> idx = find(diff(y >= LineValue))
idx =
48 52
You can then calculate the x positions of the intersection points using weighted averages (i.e. linear interpolation):
>> x2 = x(idx) + (LineValue - y(idx)) .* (x(idx+1) - x(idx)) ./ (y(idx+1) - y(idx))
x2 =
-4.24568579887939 4.28720287203057
Plot these up to verify the results:
>> figure;
>> plot(x, y, 'b.-', x2, LineValue, 'go', [x(1) x(end)], LineValue*[1 1], 'k:');
The advantages of this approach are:
The determination of the intersection points is vectorized so will work regardless of the number of intersection points.
Determining the intersection points arithmetically is presumably faster than using fzero.
Example solution using FZERO:
%# data resembling your curve
x = linspace(-100,100,100);
f = #(x) 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
VALUE = 0.8;
%# solve f(x)=VALUE
z1 = fzero(#(x)f(x)-VALUE, -10); %# find solution near x=-10
z2 = fzero(#(x)f(x)-VALUE, 10); %# find solution near x=+10
%# plot
plot(x,f(x),'b.-'), hold on
plot(z1, VALUE, 'go', z2, VALUE, 'go')
line(xlim(), [VALUE VALUE], 'Color',[0.4 0.4 0.4], 'LineStyle',':')
hold off
Are the step sizes in your data series the same?
Is the governing equation assumed to be cubic, sinuisoidal, etc..?
doc interpl
Find the zero crossings
I have accelerometer values for the 3 axes (usually when there is only gravity contains data between -1.0 and 1.0 ):
float Rx;
float Ry;
float Rz;
I make some calculations, then I get the angles for each axis.
float R = sqrt(pow(Rx,2)+pow(Ry,2)+pow(Rz,2));
float Arx = acos(Rx/R)*180/M_PI;
float Ary = acos(Ry/R)*180/M_PI;
float Arz = acos(Rz/R)*180/M_PI;
Then I set the values for the box angles in opengl
rquad = Arx;
yquad = Ary;
Which rotates my box:
glRotatef(yquad,1.0f,0.0f,0.0f);
glRotatef(rquad,0.0f,1.0f,0.0f);
It works on a hemisphere. I would like to use the full sphere and I know that I have to use the Arz value to make it work, but I don't know how can I use that for this rotation. Could you help me?
Update:
The final answer is in my case:
rquad = -atan2(Rx/R, Rz/R)*180/M_PI;
yquad = -atan2(Ry/R, Rz/R)*180/M_PI;
The correct answer is:
Roll = atan2(Y, Z) * 180/M_PI;
Pitch = atan2(-X, sqrt(Y*Y + Z*Z)) * 180/M_PI;
Source: https://www.nxp.com/docs/en/application-note/AN3461.pdf (page 10, Eqn. 25 & 26)
uesp's answer is wrong. It looks like an acceptable approximation until pitch and roll both go above 45 degrees.
I may be assuming a different orientation convention, but even if you swap axes and invert values in any consistent way, uesp's computations will never be equivalent.
While matteo's answer is correct, it does not provide the full, complete solution:
The formulas are correct:
Roll = atan2(Y, Z) * 180/M_PI;
Pitch = atan2(-X, sqrt(Y*Y + Z*Z)) * 180/M_PI;
However, when pitch is +90/-90 degrees and the X axis is vertical pointing up/down, the ideal accelerometer normalized output should be:
accX = -1 / accX = 1
accY = 0
accZ = 0
Which means a roll angle of 0 degrees; correct.
But in practice, the accelerometer output is noisy and you would get something closer to:
accX = -1 / accX = 1
accY = 0.003
accZ = 0.004
This might seem small but it will cause the roll angle to be ~30 dregrees which is not correct.
The obvious instinct would be to filter out the last digits, but this would affect precision, which is not always acceptable.
The compromise, which is very well explained in the reference app note, is to include a very small percentage of the accelerometer X axis reading in the formula for roll:
Roll = atan2( Y, sign* sqrt(Z*Z+ miu*X*X));
sign = 1 if accZ>0, -1 otherwise
miu = 0.001
The error introduced this way is dramatically smaller than the previous case: 2-3 degrees when measuring roll under the same conditions explained above.
I've tried the recommended solution (matteo's), and while it seemed to work great at first, I noticed that when the pitch approaches 90 degrees (starting at around 70 degrees but not necessarily consistent across different phones), the roll suddenly surges. When the pitch is at 90 the roll that should be around 0 is now at over 100 and keeps increasing to 180. I'm trying to think of a way to mathematically prevent this, if I restrict the roll to +90/-90 it behaves normally but I don't get the range I want (+180/-180):
Math.atan2(y, Math.sqrt((xx) + (zz))) * (180/Math.PI))
For roll, I have found you can use arctan(y/sqrt(X*X)+(z*z)) this will give roll -90/90 which is aviation standard without giving the pitch issue
I use the following calculations to convert our accelerometer readings into roll and pitch values:
Roll = atan2( sqrt(Y*Y + X*X), Z) * 180/M_PI;
Pitch = atan2( sqrt(X*X + Z*Z), Y) * 180/M_PI;
You may need to swap the X/Y/Z values or translate the Roll/Pitch depending on how your accelerometers are defined. To use them in the display them it is a simple matter of:
glRotatef (Pitch, 0.0f, 0.0f, 1.0f);
glRotatef (Roll, 1.0f, 0.0f, 0.0f);
I want to move something a set distance. However in my system there is inertia/drag/negative accelaration. I'm using a simple calculation like this for it:
v = oldV + ((targetV - oldV) * inertia)
Applying that over a number of frames makes the movement 'ramp up' or decay, eg:
v = 10 + ((0 - 10) * 0.25) = 7.5 // velocity changes from 10 to 7.5 this frame
So I know the distance I want to travel and the acceleration, but not the initial velocity that will get me there. Maybe a better explanation is I want to know how hard to hit a billiard ball so that it stops on a certain point.
I've been looking at Equations of motion (http://en.wikipedia.org/wiki/Equations_of_motion) but can't work out what the correct one for my problem is...
Any ideas? Thanks - I am from a design not science background.
Update: Fiirhok has a solution with a fixed acceleration value; HTML+jQuery demo:
http://pastebin.com/ekDwCYvj
Is there any way to do this with a fractional value or an easing function? The benefit of that in my experience is that fixed acceleration and frame based animation sometimes overshoots the final point and needs to be forced, creating a slight snapping glitch.
This is a simple kinematics problem.
At some time t, the velocity (v) of an object under constant acceleration is described by:
v = v0 + at
Where v0 is the initial velocity and a is the acceleration. In your case, the final velocity is zero (the object is stopped) so we can solve for t:
t = -v0/a
To find the total difference traveled, we take the integral of the velocity (the first equation) over time. I haven't done an integral in years, but I'm pretty sure this one works out to:
d = v0t + 1/2 * at^2
We can substitute in the equation for t we developed ealier:
d = v0^2/a + 1/2 * v0^2 / a
And the solve for v0:
v0 = sqrt(-2ad)
Or, in a more programming-language format:
initialVelocity = sqrt( -2 * acceleration * distance );
The acceleration in this case is negative (the object is slowing down), and I'm assuming that it's constant, otherwise this gets more complicated.
If you want to use this inside a loop with a finite number of steps, you'll need to be a little careful. Each iteration of the loop represents a period of time. The object will move an amount equal to the average velocity times the length of time. A sample loop with the length of time of an iteration equal to 1 would look something like this:
position = 0;
currentVelocity = initialVelocity;
while( currentVelocity > 0 )
{
averageVelocity = currentVelocity + (acceleration / 2);
position = position + averageVelocity;
currentVelocity += acceleration;
}
If you want to move a set distance, use the following:
Distance travelled is just the integral of velocity with respect to time. You need to integrate your expression with respect to time with limits [v, 0] and this will give you an expression for distance in terms of v (initial velocity).
I have not marked this question Answered yet.
The current accepted answer got accepted automatically because of the Bounty Time-Limit
With reference to this programming game I am currently building.
As you can see from the above link, I am currently building a game in where user-programmable robots fight autonomously in an arena.
Now, I need a way to detect if a robot has detected another robot in a particular angle (depending on where the turret may be facing):
alt text http://img21.imageshack.us/img21/7839/robotdetectionrg5.jpg
As you can see from the above image, I have drawn a kind of point-of-view of a tank in which I now need to emulate in my game, as to check each point in it to see if another robot is in view.
The bots are just canvases that are constantly translating on the Battle Arena (another canvas).
I know the heading the turret (the way it will be currently facing), and with that, I need to find if there are any bots in its path(and the path should be defined in kind of 'viewpoint' manner, depicted in the image above in the form of the red 'triangle'. I hope the image makes things more clear to what I am trying to convey.
I hope that someone can guide me to what math is involved in achieving this problem.
[UPDATE]
I have tried the calculations that you have told me, but it's not working properly, since as you can see from the image, bot1 shouldn't be able to see Bot2 . Here is an example :
alt text http://img12.imageshack.us/img12/7416/examplebattle2.png
In the above scenario, Bot 1 is checking if he can see Bot 2. Here are the details (according to Waylon Flinn's answer):
angleOfSight = 0.69813170079773179 //in radians (40 degrees)
orientation = 3.3 //Bot1's current heading (191 degrees)
x1 = 518 //Bot1's Center X
y1 = 277 //Bot1's Center Y
x2 = 276 //Bot2's Center X
y2 = 308 //Bot2's Center Y
cx = x2 - x1 = 276 - 518 = -242
cy = y2 - y1 = 308 - 277 = 31
azimuth = Math.Atan2(cy, cx) = 3.0141873380511295
canHit = (azimuth < orientation + angleOfSight/2) && (azimuth > orientation - angleOfSight/2)
= (3.0141873380511295 < 3.3 + 0.349065850398865895) && (3.0141873380511295 > 3.3 - 0.349065850398865895)
= true
According to the above calculations, Bot1 can see Bot2, but as you can see from the image, that is not possible, since they are facing different directions.
What am I doing wrong in the above calculations?
The angle between the robots is arctan(x-distance, y-distance) (most platforms provide this 2-argument arctan that does the angle adjustment for you. You then just have to check whether this angle is less than some number away from the current heading.
Edit 2020: Here's a much more complete analysis based on the updated example code in the question and a now-deleted imageshack image.
Atan2: The key function you need to find an angle between two points is atan2. This takes a Y-coordinate and X-coordinate of a vector and returns the angle between that vector and the positive X axis. The value will always be wrapped to lie between -Pi and Pi.
Heading vs Orientation: atan2, and in general all your math functions, work in the "mathematical standard coordinate system", which means an angle of "0" corresponds to directly east, and angles increase counterclockwise. Thus, an "mathematical angle" of Pi / 2 as given by atan2(1, 0) means an orientation of "90 degrees counterclockwise from due east", which matches the point (x=0, y=1). "Heading" is a navigational idea that expresses orientation is a clockwise angle from due north.
Analysis: In the now-deleted imageshack image, your "heading" of 191 degrees corresponded to a south-south-west direction. This actually an trigonometric "orientation" of -101 degrees, or -1.76. The first issue in the updated code is therefore conflating "heading" and "orientation". you can get the latter from the former by orientation_degrees = 90 - heading_degrees or orientation_radians = Math.PI / 2 - heading_radians, or alternatively you could specify input orientations in the mathematical coordinate system rather than the nautical heading coordinate system.
Checking that an angle lies between two others: Checking that an vector lies between two other vectors is not as simple as checking that the numeric angle value is between, because of the way the angles wrap at Pi/-Pi.
Analysis: in your example, the orientation is 3.3, the right edge of view is orientation 2.95, the left edge of view is 3.65. The calculated azimith is 3.0141873380511295, which happens to be correct (it does lie between). However, this would fail for azimuth values like -3, which should be calculated as "hit". See Calculating if an angle is between two angles for solutions.
Calculate the relative angle and distance of each robot relative to the current one. If the angle is within some threshold of the current heading and within the max view range, then it can see it.
The only tricky thing will be handling the boundary case where the angle goes from 2pi radians to 0.
Something like this within your bot's class (C# code):
/// <summary>
/// Check to see if another bot is visible from this bot's point of view.
/// </summary>
/// <param name="other">The other bot to look for.</param>
/// <returns>True iff <paramref name="other"/> is visible for this bot with the current turret angle.</returns>
private bool Sees(Bot other)
{
// Get the actual angle of the tangent between the bots.
var actualAngle = Math.Atan2(this.X - other.X, this.Y - other.Y) * 180/Math.PI + 360;
// Compare that angle to a the turret angle +/- the field of vision.
var minVisibleAngle = (actualAngle - (FOV_ANGLE / 2) + 360);
var maxVisibleAngle = (actualAngle + (FOV_ANGLE / 2) + 360);
if (this.TurretAngle >= minVisibleAngle && this.TurretAngle <= maxVisibleAngle)
{
return true;
}
return false;
}
Notes:
The +360's are there to force any negative angles to their corresponding positive values and to shift the boundary case of angle 0 to somewhere easier to range test.
This might be doable using only radian angles but I think they're dirty and hard to read :/
See the Math.Atan2 documentation for more details.
I highly recommend looking into the XNA Framework, as it's created with game design in mind. However, it doesn't use WPF.
This assumes that:
there are no obstacles to obstruct the view
Bot class has X and Y properties
The X and Y properties are at the center of the bot.
Bot class a TurretAngle property which denotes the turret's positive angle relative to the x-axis, counterclockwise.
Bot class has a static const angle called FOV_ANGLE denoting the turret's field of vision.
Disclaimer: This is not tested or even checked to compile, adapt it as necessary.
A couple of suggestions after implementing something similar (a long time ago!):
The following assumes that you are looping through all bots on the battlefield (not a particularly nice practice, but quick and easy to get something working!)
1) Its a lot easier to check if a bot is in range then if it can currently be seen within the FOV e.g.
int range = Math.sqrt( Math.abs(my.Location.X - bots.Location.X)^2 +
Math.abs(my.Location.Y - bots.Location.Y)^2 );
if (range < maxRange)
{
// check for FOV
}
This ensures that it can potentially short-cuircuit a lot of FOV checking and speed up the process of running the simulation. As a caveat, you could have some randomness here to make it more interesting, such that after a certain distance the chance to see is linearly proportional to the range of the bot.
2) This article seems to have the FOV calculation stuff on it.
3) As an AI graduate ... nave you tried Neural Networks, you could train them to recognise whether or not a robot is in range and a valid target. This would negate any horribly complex and convoluted maths! You could have a multi layer perceptron [1], [2] feed in the bots co-ordinates and the targets cordinates and recieve a nice fire/no-fire decision at the end. WARNING: I feel obliged to tell you that this methodology is not the easiest to achieve and can be horribly frustrating when it goes wrong. Due to the (simle) non-deterministic nature of this form of algorithm, debugging can be a pain. Plus you will need some form of learning either Back Propogation (with training cases) or a Genetic Algorithm (another complex process to perfect)! Given the choice I would use Number 3, but its no for everyone!
It can be quite easily achieved with the use of a concept in vector math called dot product.
http://en.wikipedia.org/wiki/Dot_product
It may look intimidating, but it's not that bad. This is the most correct way to deal with your FOV issue, and the beauty is that the same math works whether you are dealing with 2D or 3D (that's when you know the solution is correct).
(NOTE: If anything is not clear, just ask in the comment section and I will fill in the missing links.)
Steps:
1) You need two vectors, one is the heading vector of the main tank. Another vector you need is derived from the position of the tank in question and the main tank.
For our discussion, let's assume the heading vector for main tank is (ax, ay) and vector between main tank's position and target tank is (bx, by). For example, if main tank is at location (20, 30) and target tank is at (45, 62), then vector b = (45 - 20, 62 - 30) = (25, 32).
Again, for purpose of discussion, let's assume main tank's heading vector is (3,4).
The main goal here is to find the angle between these two vectors, and dot product can help you get that.
2) Dot product is defined as
a * b = |a||b| cos(angle)
read as a (dot product) b since a and b are not numbers, they are vectors.
3) or expressed another way (after some algebraic manipulation):
angle = acos((a * b) / |a||b|)
angle is the angle between the two vectors a and b, so this info alone can tell you whether one tank can see another or not.
|a| is the magnitude of the vector a, which according to the Pythagoras Theorem, is just sqrt(ax * ax + ay * ay), same goes for |b|.
Now the question comes, how do you find out a * b (a dot product b) in order to find the angle.
4) Here comes the rescue. Turns out that dot product can also be expressed as below:
a * b = ax * bx + ay * by
So angle = acos((ax * bx + ay * by) / |a||b|)
If the angle is less than half of your FOV, then the tank in question is in view. Otherwise it's not.
So using the example numbers above:
Based on our example numbers:
a = (3, 4)
b = (25, 32)
|a| = sqrt(3 * 3 + 4 * 4)
|b| = sqrt(25 * 25 + 32 * 32)
angle = acos((20 * 25 + 30 * 32) /|a||b|
(Be sure to convert the resulting angle to degree or radian as appropriate before comparing it to your FOV)
This will tell you if the center of canvas2 can be hit by canvas1. If you want to account for the width of canvas2 it gets a little more complicated. In a nutshell, you would have to do two checks, one for each of the relevant corners of canvas2, instead of one check on the center.
/// assumming canvas1 is firing on canvas2
// positions of canvas1 and canvas2, respectively
// (you're probably tracking these in your Tank objects)
int x1, y1, x2, y2;
// orientation of canvas1 (angle)
// (you're probably tracking this in your Tank objects, too)
double orientation;
// angle available for firing
// (ditto, Tank object)
double angleOfSight;
// vector from canvas1 to canvas2
int cx, cy;
// angle of vector between canvas1 and canvas2
double azimuth;
// can canvas1 hit the center of canvas2?
bool canHit;
// find the vector from canvas1 to canvas2
cx = x2 - x1;
cy = y2 - y1;
// calculate the angle of the vector
azimuth = Math.Atan2(cy, cx);
// correct for Atan range (-pi, pi)
if(azimuth < 0) azimuth += 2*Math.PI;
// determine if canvas1 can hit canvas2
// can eliminate the and (&&) with Math.Abs but this seems more instructive
canHit = (azimuth < orientation + angleOfSight) &&
(azimuth > orientation - angleOfSight);
Looking at both of your questions I'm thinking you can solve this problem using the math provided, you then have to solve many other issues around collision detection, firing bullets etc. These are non trivial to solve, especially if your bots aren't square. I'd recommend looking at physics engines - farseer on codeplex is a good WPF example, but this makes it into a project way bigger than a high school dev task.
Best advice I got for high marks, do something simple really well, don't part deliver something brilliant.
Does your turret really have that wide of a firing pattern? The path a bullet takes would be a straight line and it would not get bigger as it travels. You should have a simple vector in the direction of the the turret representing the turrets kill zone. Each tank would have a bounding circle representing their vulnerable area. Then you can proceed the way they do with ray tracing. A simple ray / circle intersection. Look at section 3 of the document Intersection of Linear and Circular Components in 2D.
Your updated problem seems to come from different "zero" directions of orientation and azimuth: an orientation of 0 seems to mean "straight up", but an azimuth of 0 "straight right".