You will need
- - a sheet of paper;
- - gon;
- a pair of compasses;
- - pattern;
- - eraser;
Mark on the drawing point which is the center of the spiral of Archimedes. Mark the centre of the letter O.
Build from the center of the spiral, the circle whose radius is equal to the pitch of the helix. The pitch of the spiral of Archimedes equal to the distance travelled by the point on the surface of the circle for one complete revolution.
In descriptive geometry, the spiral of Archimedes relates to a radial curve. It is constructed using curves connecting the points on the circle. To get construction points, divide the circle into several equal parts with straight lines. For example, 8.
Number for the convenience of straight lines, dividing the circumference in the direction of rotation of the circle.
Divide the radius of the constructed circle to the number to which the divided circle using straight lines. With the help of a compass or ruler, divide the last in the numbering straight on the value of the marks. You only need to divide the interval between the center of the circle O and the point of intersection of a straight circle.
Number the marks, starting with the closest to the center of the circle. You can use numbers or letters in alphabetical order.
Using the compass, draw an arc of a circle center O. Arc starts from a straight line, which separated the marks and is conducted to a straight line under the number 1. Label the point where the arc is connected to the line 1 of figure 1. Similarly, construct the following arc from the marked straight line to straight at number 2. Designate a point of connection with figure 2, and then mark that point on all straight separates the circle.
Using the patterns, connect the center circle with the first point. Then connect the first point with the second and so connect all the marked points. You will get the first turn of the spiral of Archimedes. Try to connect the dots as evenly as possible. To get the spiral of Archimedes, higher accuracy, divide the circle into a greater number of equal parts and construct the corresponding number of arcs.
Advice 2: How to determine the type of conic curve
The answer is quite simple. Transform the General equation of a curve of second order to the canonical form. The desired curves are only three, and it is an ellipse, hyperbola and parabola. The form of the relevant equations can be seen in additional sources. There you can verify that the full procedure of reduction to canonical form should be avoided because of its bulkiness.
Question about clarification of a curve of the second order is more qualitative than quantitative task. In the most General case, the solution may start with the given equation of the line of second order (see Fig. 1). In this equation all the coefficients are some constants. If you have forgotten the equation of ellipse, parabola and hyperbola in canonical form, view their additional sources to this article or any textbook.
Compare the General equation with each of those canonical. It is easy to conclude that if the coefficients a≠0, C≠0, and their sign are the same, after any transformation leading to the canonical form, the resulting ellipse. If the sign is hyperbole. The parabola will meet the situation, when the coefficients or A or C (but not both) is zero. Thus, the response is received. Only here the numerical characteristics no, except for those coefficients that are specific to the problem statement.
There is another way of obtaining answer to a question. This application of the General polar equation of curves of second order. This means that in polar coordinates, all three fit into the Canon of the curves (Cartesian coordinates) are written almost the same equation. And although it is in the Canon and does not fit – is possible here the list of curves of the second order to expand indefinitely (aplicata Bernoulli, Lissajous figure, etc.).
We restrict the ellipse (in General) and hyperbole. Parabola occurs automatically, as the intermediate case. The fact that the original ellipse is defined as locus of points for which the sum of the focal radii r1+r2=2a =const. For hyperbole |r1-r2|=2a=const. Put the foci of the ellipse (hyperbola) F1(-c, 0), F2(c, 0). Then the focal radii of the ellipse are equal (see Fig. 2A). For the right branch of the hyperbola, see figure 2b.
Polar coordinates ρ=ρ(φ) should be entered using the focus as the polar center. Then we can put ρ=r2 and after some transformations will get to the right areas of the ellipse and the parabola, the polar equations (see Fig. 3). While a is the semimajor axis of the ellipse (the imaginary for the hyperbola) is the x coordinate of the focus, about the parameter b in the figure.
Given the formulas of figure 2, the value ε is called the eccentricity. From the formulas of figure 3 it follows that all other values with it as in any way involved. Indeed, since ε is connected to all the main curves of the second order, based on it and can make major decisions. Namely, if ε1 is hyperbole. ε=1 – parabola. It has a deeper meaning. In a much extremely difficult course "equations of mathematical physics" classification of differential equations is made on the same basis.