trigonometry

问题 I have a method in my Enemy class called huntPlayer . It takes a player object p . Here it is: def huntPlayer(self, p): if self.dist2p < 200: self.hunting = True if p.x > self.rect.x: self.rect.x += self.speed #this is a constant at value 1 elif p.x < self.rect.x: self.rect.x -= self.speed else: self.rect.x += 0 if p.y > self.rect.y: self.rect.y += self.speed elif p.y < self.rect.y: self.rect.y -= self.speed else: self.rect.y += 0 else: self.rect.x += 0 self.rect.y += 0 The enemies are

问题 I have an excel formula that I need to put into an MS-Access query. Unfortunately, an Arc Cosine function does not seem to be part of Access's repertoire. So in Excel, part of the formula is ACOS(Lots of stuff). So my question is how can I achieve an ACOS() Excelfunction in MS-Access? Note: In case anyone is wondering or in case it actually matters for some odd reason, here is the full equation as it appears in Excel: =(ACOS(SIN(RADIANS(AX2))*SIN(RADIANS(AZ2))+COS(RADIANS(AX2))*COS(RADIANS

问题 I have been doing some research as to how sine and cosine can be calculated. I found a couple of "standard" methods, including a lookup table, a CORDIC algorithm, and Taylor series. I also found that most modern processors have an assembler instruction calculating trigonometric functions. What I want to know is how those commands work. So, my question is: What specific algorithm do current gen processors use for calculating sine and cosine? 回答1: The answer to a related, but different question

问题 I have an ellipse-shaped distribution that I think is a conic section. I want to rotate the points so that the distribution becomes circular-shaped, as if I was looking at the conic section directly over the top of the cone. Here's some example data (generated using the function here) X_df <- structure(list(x = c(550.685479223573, 411.808342674744, 125.337513241526, -46.6813176776531, 54.1090479024869, 335.045593380922, 538.806846993829, 476.123346783785, 207.359201714354, -23.3704356149293,

问题 I have an assignment where I need to approximate Pi in a computationally efficient manner. Here is my strategy: I use a unit circle, the angle bisector of an isoceles triangle, and the definition of sin. I drew a diagram: For example, if I want to use an hexagon (6 points/6 sides), I simply need to compute a :( 0.5*sin(2*pi/2*x ) and multiply it by ( 2*x ). Finally, since Pi = Circumference/Diameter , then my approximation of Pi = polygon perimeter (since Diameter = 1 ). Essentially: from

问题 Here's my code: <?php $lat1 = 35.697959; $lat2 = 35.697959; $lon1 = 139.707085; $lon2 = 139.707085; $theta = $lon1 - $lon2; $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta)); var_dump($dist); // returns 1 $dist = acos($dist); var_dump($dist); // returns NAN As you can see from the comments, $dist is equal to 1 after the calculation, but when I apply acos() it returns NAN . If I try to replicate it manually, it works just fine:

问题 How do I generate a simple sine wave in matlab? I would like to generate a wave which represents a temperature signal with an amplitude of 15 degrees during a 24 hour period, how can I do this? t = 1:24 x = 15.*sin(pi*t) plot(t,x) where 15 is the amplitude. This does not generate a sine wave as I expected. I was expecting to see one wave which extends over a 24 hour period with an amplitude of 15, say with the lowest value of 5 and a maximum of 20 (how do I include these in the equation?).

问题 In C library math.h , there was a sincos function which was pretty efficient, because it computed both sine and cosine in a time closer to a single call to sin() or cos() than to the total time of calling both. Is there such function in C++ standard library? 回答1: Is there no such function in c++ standard library? No, unfortunately there isn't. In C library math.h, there was a sincos function On Linux, it is available as GNU Extension. It's not standard in C either. 回答2: Just use sin and cos

问题 I need to find a point where a line (its origin is ellipse' center) intersects an ellipse in 2D... I can easily find a point on a circle, because I know an angle F and the circle' radius (R): x = x0 + R * cosF y = y0 + R * sinF However I just can't figure how am I supposed to deal with an ellipse... I know it's dimensions (A & B), but what is the way of finding parameter T?! x = x0 + A * cosT y = y0 + B * sinT From what I understand the parameter T (T angle) is not far from the F angle

问题 I'm creating a solar system animation using canvas and have a problem calculating an object's position (x,y) values. Earth is orbiting around the Sun, and I need to calculate and update the Earth's position at each frame. And using Earth's position, I'll orbit the Moon around the Earth. The related function is this: orbitAround : function (master) { master.makeOrbitCenter(); this.increaseOrbitAngle(); context.rotate(this.orbit.angle * Math.PI/180); context.translate(this.orbit.distance, 0); /