[英]Triangle class using an abstract shape class
So I have this triangle class I need to create using an abstract class. 因此,我需要使用一个抽象类来创建这个三角形类。 It will also be drawn by a tester class.
它也将由测试人员类绘制。 I am part of the way through it but I am having serious issues with the math portion.
我是其中的一部分,但是数学部分存在严重问题。 I have set the coordinates in the tester class, I have no idea of how to get the pen to turn a certain degree to draw the next side of the triangle.
我已经在测试器类中设置了坐标,我不知道如何使笔旋转一定程度以绘制三角形的下一个侧面。 Attached is all the classes and I have so far.
附件是所有课程,到目前为止,我已经掌握了。 Any help will be appreciated.
任何帮助将不胜感激。
Tester class 测试员班
import TurtleGraphics.*;
public class TestShapes1 {
public static void main (String[] args) {
// Declare and instantiate a pen, a circle and a wheel
Pen p = new StandardPen();
//Shape s1 = new Circle1 (20, 20, 20);
//Shape s2 = new Wheel1 (-20, -20, 20, 6);
Shape1 t2 = new Triangle1 (0, 0, 50, 0, 0, 30);
// Draw the circle and wheel
//s1.draw (p);
t2.draw (p);
}
}
Shape Class 形状等级
import TurtleGraphics.Pen;
public interface Shape1 {
public double area();
public void draw (Pen p);
public double getXPos();
public double getYPos();
public void move (double xLoc, double yLoc);
public void stretchBy (double factor);
public String toString();
}
Triangle Class 三角类
import TurtleGraphics.Pen;
public class Triangle1 implements Shape1 {
private double x1, y1, x2, y2, x3, y3;
private double s1, s2, s3;
private double d1, d2;
//private double height, width;
public Triangle1() {
x1 = 0;
y1 = 0;
x2 = 1;
y2 = 0;
x3 = 0;
y3 = 1;
//height = 1;
//width = 1;
}
public Triangle1 (double xLoc1, double yLoc1, double xLoc2, double yLoc2, double xLoc3, double yLoc3) {
x1 = xLoc1;
y1 = yLoc1;
x2 = xLoc2;
y2 = yLoc2;
x3 = xLoc3;
y3 = yLoc3;
//height = h;
//width = w;
}
public double area() {
return (Math.abs(x1*y2-x2*y1+x2*y3-x3*y2+x3*y1-x1*y3))/2.0;
}
public void draw (Pen p) {
s1 = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
s2 = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
s3 = Math.sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
p.up();
p.move (x1, y1);
p.down();
p.setDirection (0);
p.move (s1);
d1 = (Math.acos((s2*s2+s3*s3-s1*s1)/(2.0*s2*s3)))*180/Math.PI;
p.turn (180 - d1);
p.move (s2);
d2 = (Math.acos((s3*s3+s1*s1-s2*s2)/(2.0*s3*s1)))*180/Math.PI;
p.turn (180 - d2);
p.move (s3);
p.turn (-90);
//p.move ();
}
public double getXPos() {
return x1;
}
public double getYPos() {
return y1;
}
public void move (double xLoc, double yLoc) {
x1 = x1 + xLoc;
y1 = y1 + yLoc;
x2 = x2 + xLoc;
y2 = y2 + yLoc;
x3 = x3 + xLoc;
y3 = y3 + yLoc;
}
public void stretchBy (double factor) {
x1 *= factor;
y1 *= factor;
}
public String toString() {
String str = "TRIANGLE\n";
// + "Width & Height: " + width + " & " + height +"\n"
// + "(X,Y) Position: (" + xPos + "," + yPos + ")\n"
// + "Area: " + area();
return str;
}
}
You don't need any math. 您不需要任何数学。 Just pass the degrees to
p.turn()
. 只需将度数传递给
p.turn()
。 So use 所以用
p.turn(180);
instead of 代替
d1 = (Math.acos((s2*s2+s3*s3-s1*s1)/(2.0*s2*s3)))*180/Math.PI;
p.turn (180 - d1);
See the documentation for reference: 请参阅文档以供参考:
The degrees can be an integer or floating-point number.
度可以是整数或浮点数。 Example: pen.turn(-45);
示例:pen.turn(-45); Rotate the pen 45 degrees clockwise.
将笔顺时针旋转45度。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.