If the tracking axis r is tilted toward the south as is often done, the equations above simplify on setting to degrees. Offset Aperture In some tracking designs, the collector aperture is offset relative to the tracking axis by an angle as shown in Figure 4. For this design, the tracking angle is still as computed by Equation 4. However, the angle of incidence is no longer that computed by Equation 4.
The concept of the moment of a force comes from the law of the lever, discovered by Archimedes. It corresponds to the torque exerted on a lever by a force. A lever consists of a rigid bar which is free to turn about a fixed point called the fulcrum.
Moment of a force. The torque produced by a force F is also defined as Fd. The two terms are synonymous.
Forces F1 and F2 produce a counterclockwise torque and forces F3, F4 and F5 produce a clockwise torque. The law of the lever states that in order to have equilibrium the sum of the counterclockwise torques or moments must be equal to the sum of the clockwise torques. If the sums are not equal there will be rotation.
Computing centers of gravity. We wish now to deal with the problem of computing the location of the center of gravity, or center of mass, of a body. The center of gravity is the same as the center of mass since weight and mass are proportional. However, in developing the ideas involved we need to assume a gravitational field and will speak of the center of gravity.
In developing the ideas for computing the location of the center of gravity we will view a body as an assemblage of individual particles.
The earth exerts an attraction on each individual particle of a body and the weight of a body is the sum total of all the forces on all the particles making up the body.
Thus we will consider the problem of finding the location of the center of gravity for assemblages of particles in space. Consider a steel rod resting on a pivot as shown in Fig.
If the pivot is directly below the center of gravity, the rod is balanced, and the sum of all the clockwise moments from particles to the right of the pivot is equal to the sum of all the counterclockwise moments from the particles to the left of the pivot.
The upward force F exerted by the pivot on the rod, as shown in Fig. Now let us consider the situation in which the pivot is at the left end of the rod as shown in Fig. Suppose an upward force F is exerted directly below the center of gravity and equal in magnitude to the weight of the rod.
Then the rod will be balanced, no force will be exerted on the pivot, and the sum of all of the clockwise moments from the particles of the rod will be equal to the counterclockwise moment Fd produced by force F where d is the distance from the pivot to the center of gravity, as shown in the figure.
Thus if we were able to compute the clockwise moments of the particles of the rod with respect to the pivot point and knew the weight of the rod, we would then be able to compute the distance d from the pivot to the center of gravity. Consider now a system of n point masses situated along a horizontal line, as shown in Fig.
Suppose the weights of the masses are w1, w2, w3, Let M1 be the sum of the clockwise moments of the n masses with respect to the pivot point. Then the counterclockwise moment M2 due to F is given by Fd where d is the distance from the pivot to the center of gravity.
We lay an x-axis of a coordinate system along the rod with the origin at the pivot. We consider the rod to consist of an assemblage of particles m1, m2, The weight W of the rod is given by Since mass is proportional to weight, 3 can also be expressed in terms of mass.
In terms of mass it is where x is the distance of the infinitesimal element of mass dm from the pivot. We have considered a one-dimensional case where masses are distributed out in a line, as in a rod. The ideas are easily extended to assemblages of points in two and three-dimensional space.
Consider a system of n point masses m1, m2, Again the point masses are conceived of as being attached to a rigid weightless framework. The masses are located at points x1, y1x2, y2x3, y3The moment of mass m1 with respect to the y-axis is x1w1.
Its moment with respect to the x-axis is y1w1. The x and y coordinates of the center of gravity of this collection of point masses is then where W is the combined weight of the n point masses. For an assemblage of point masses in three-dimensional space as shown in Fig.In vertex form, a quadratic function is written as y = a(x-h) 2 + k See also Quadratic Explorer - standard form.
In the applet below, move the sliders on the right to change the values of a, h and k and note the effects it has on the graph. May 18, · ANSWER: The vertex is point (-2, -1) and x = -2 is the equation of the axis of symmetry.
*The x-coordinate is opposite of what it appears to be in the equation (it is -2 instead of 2) because, if you look at the original formula, you are subtracting h, not adding arteensevilla.com: Resolved. You will also need to work the other way, going from the properties of the parabola to its equation.
Write an equation for the parabola with focus at (0, –2) and directrix x = 2.; The vertex is always halfway between the focus and the directrix, and the parabola always curves away from the directrix, so I'll do a quick graph showing the focus, the directrix, .
Symmetry Having symmetry (or being symmetric) means that something is a mirror image over a line in space.. Look at the picture on the right The left halfis a perfect mirror image of the right half over that yellow line.. (Ignoring the .
On this page, we will practice drawing the axis on a graph, learning the formula, stating the equation of the axis of symmetry when we know the parabola's equation Explore how the graph and equation relate to the axis of symmetry, by using our interactive program below.
Pranjal Srivastava, To test for symmetry algebraically about the y axis you take the equation y = f(x) and substitute -x for x and see whether you get the same equation .