Elmann Creative Corner Fandom TV and computer Bents Space.

Section Five: Chapter 1. Teaching Arithmetic: One of the most facinating studies in the world is mathmatics, = substantive Middle English = 162+162=324 ( orig. original(ly a. OFr. Old French arismetique, for Latin arithmetica. The science of numbers; the art of computation by figures Arithmetical knowledge, computation -90. A treatise on computation -74 -90 = -164 + 324 = 160. These roguish Arsmetrique gibbets or flesh-hookes, and cyphers or round 009. But now tis odders beyond Arithmetick Cor. Coriolanus 111. i. 245 = 3 x 9 = 27 + 245 = 272 +- 160 = 112 and this using follows A concave spherical reflector is 12 in. in diameter and the concavity is 1 in. deep. Calculate the area inside this reflector and find out how it compares with the area of a flat disk of the same diameter. To find the diameter d of the whole sphere of which the reflector surface forms a part, we have A spherical buoy 4 ft. in diameter is made of material weighing 0.05 lb. per sq. in. What is the finished weight? Surface area of buoy = 4 x area of great circle = 4 x phir 2 = 4 x phi x 24 2 = 7240 sq. in. Weight of buoy = 0.05 x 7240 = 362 lb. and 6 x 6 = 1 (d-1) = d - 1 and d = 36 + 1 = 37. Area of reflector = area of 1 in., belt of cylinder 37 in. in diameter = phi x 37 x 1 = 116 sq.in. Area of 12 in. disk = phi/4 x 144 = 112.5 sq in. and Area of reflector / Area of disk = 1.031 x 362 lb. 373.222 +- 324 = 49.222 x 0.25 = 12.3055 = Work and Energy  and here The component of the 50-lb. weight perpendicular to the slope now concerns us; it is root 50 2 - 10 2 = 49 lb. Friction resistance to sliding = 0.25 x 49 = 12 1/4 lb. Work done in overcoming this = 12 1/4 x 100 = 1225 ftg-lb. Therefore the total work done in dragging the weight = 1000 + 1225 = 2225 ft-lb. Energy. When a body is able to do work it is said to possess ``energy ``, this being defined as the capacity for doing work. Thus, a 10-lb. weight held in some way at a height of 20 ft. above the ground is able, on release, to do 10 x 20 = 200 ft.-lb. of work. It could lift up other weights in its descent, or it could be connected by strings in such a way as to pull a truck up a slope. The energy of the suspendel weight is this 200 ft.-lb. of work that it is capable of doing by virtue of its position. Such energy as bodies possess by virtue of position is called potential energy. A heavy body travelling at a high speed can also do work, as we have seen in several examples already. Thus, a travelling goods wagon bumping into one at rest will overcome its resistance and push it along. A heavy train running into buffer stops will squeeze them up against the powerful opposing forces of springs or water-pressure. This energy a body possesses by virtue of its motion is called kinetic energy. Its amount is easily calculated. Suppose that a  force  f  is used up entirely in giving an acceleration  a  to a body weighing w lb. Then we already know:  f = w/g a. If the force continues to push through a distance s, we shall have: f x s = w/g a x s.  But for uniformly accelerated motion: s = 1/2at 2 and also Uniformly Accelerated Motion. A stone or other body let fall from a great height rushes towards the ground with an ever-increasing speed. After one second of fall, its velocity is 32 f.p.s. (roughly), after 2 sec. this velocity has increased by another 32 f.p.s. to 64 f.p.s. After 3 sec. the speed has become 3 x 32 = 96 f.p.s., and this is the way it goes on growing. Motion like this, in which the velocity increases by the same amount with every second that elapses, is called uniformly accelerated motion. If the increment in velocity is a for every second which elapses the total increment in velocity will be at. If the body began with no velocity at all, its velocity v after t sec. will be given by the formula: v = at. If, on the other hand, the body began by having a velocity of u, then its final velocity v after t sec. will be given by the formula: v = u + at. A train starting from rest with an acceleration of a mile an hour per second will attain a speed of 30 m.p.h. in 30 sec. A train passing a signal-box at 30 miles an hour and having acceleration of 1/2 m.p.h. per second will have a speed of 45 m.p.h. 30 sec .after it has passed the signal-box. These are illustrations of what the above formulae signify. The velocity-time graph for a uniformly accelerated motion, starting from rest, is shown in the sketch on this page. With every second of time that elapses, the speed increases by some definite amount, so that the graph is a straight line representing the formula v = at. The distance s travelled after time t is equal to the triangular area under the graph so far as this goes in time t; and because the area of a triangle is given by the product of half the base by the height: s = 1/2 vt. If we put v = at, this formula becomes:  s = 1/2  at 2. and so: f x s = w/g a x 1/2 at 2 or: work done = w/2 g x  a 2 t 2. Again: at = v, where v  is the  final velocity attained. Consequently: work done = Kinetic Energy = wv 2/2g. We have been talking about the work put into a body to accelerate it, but, of course, this is the same in amount as the work that can be recovered from it as it slows down. That is why we can call it the kinetic energy. The amount of energy in fast-moving bodies is rather surprising. Consider, for instance, the energy in a missile weighing 1/10 lb. travelling at 1000 f.p.s. It is. K.E. = 1000 2/2 x 10 x 32 = 100,000/64 = 1562.5 ft.-lb. Could this energy be harnessed it would lift a man weighing 156 1/4 lb. through a height of 10 ft. It is not surprising that bullets and shells have the ability to force their way through anything in their path. The gun from which the above missile is fired would have the same momentum as the missile, on recoiling, and the velocity corresponding to this would be 1000/100 = 10 f.p.s. if the gun had a weight of 10 lb. or 100 times that of the bullet. The energy of recoil is : K.E. = 10 x 10 2/64 = 15.625 ft.-lb. This is a very small quantity of energy, which explains why the recoil of a gun, though an equal and opposite reaction to the projection of the missile, is not so nearly so damaging.The Conservation of Energy. When a body falls from a height it loses potential energy, or energy of position, and gains kinetic energy or energy of motion. The sum of the two remains the same throughout the descent, for energy is not destroyed. When energy seems to be destroyed (as, for instance, when friction slows down anything that is moving) it is really being converfted into some other kind of energy. Thus, the energy disappearing in friction effects reappears as heat, as you can easily find out if you will feel the brake drums of a car after it has descended a hill or been brought to a standstill quickly from a great speed. The indestructibility of energy causes the total amount of it to be always the same in any place where none can get in or get out. This is the natural law known as the Conservation of Energy. An interesting example of the Conservation of Energy is the simple pendulum. In in extreme positions to right or left, the bob comes to rest at the highest points in its travel. At these points its potential energy is at its greatest. When passing through the mid-position the bob is at its lowest point, but here it is travelling at its greatest speed because all the potential energy has been converfted into kinetic energy. A moment later it is converted back to potential energy again. During the to-and-fro motion of a pendulum there is a repeated change of potential to kinetic energy and back again. The same thing happens in other kinds of vibration, and when a weight bobs up and down on the end of a helical spring there are three kinds of energy involved. These are the potential and kinetic energies of the weight and also the energy stored in the spring as this is stretched to overcome its elasticity. In mountainous countries, where may be great volumes of water tumbling over crags and precipiices, it is posible to convert potential energy into pressure energy or kinetic energy for utilisation in one form or another of water-turbine. The method is described in as Energy of water in jet = initial energy of water. The enerfgy of the water to start with is simply h = 300, since P/w = 0 and v 2/g2 = 0. The energy of water emerging from the jet is simply v 2/2g since h= 0 and P/w = 0. v 2/2g = 300 v = root 64 x 300 = 138 ft. per sec. Area of nozzle must be equal to the area of the pipe-line divided by 138/3. Area of nozzle = 0.533 x 3 /138 = 0.01155 sq.ft. Diameter of nozzle = root 0.01155/0.7854 = 0.121 ft. = 1.452 in. Energy of jet per lb. of water passing = 300 ft.-lb. Total energy of jet = 300 x 100 = 30,000 ft.-lb. per sec. Energy utilised by Pelton wheel is 75 per cent of this, or 22,500 ft.-lb. per second. Since 1 horse-power is equivalent to 33,000 ft.-lb. per minute, or 550 ft.-lb per second. H.P. of Pelton wheel = 22500/550 = 40.9. You will see from the above example that the Pelton wheel here is really the practicable alternative to an overshot water-wheel 300 ft. in diameter. The water-wheel are splendid power generators and are highly efficient even in their older forms. There is, however, a limit to the fall that can be unilised by a simple wheel 100 or 200 ft. in diameter, though wheels 20ft. and 30 ft. in diameter are common enough. To utilise very great falls of water it is necessary to resort to the turbine type of wheel. The potential energy in the water is converted by means of pipes into pressure energy and velocity energy. Water amounting to 100 lb. a second is supplied to a Pelton wheel nozzle from a point 300 ft. highjer up a hillside. Calculate (1) the diameter of a pipe-line to bring the water down to the nozzle at a constant speed of 3 ft. per second; (2) the exit diameter of a nozzle to convert all the pressure energy of the water to kinetic energy; (3) the horse-power of the Pelton wheel if its efficiency is 75 per cent. Volume of water passing per second down the pipe is 100/62.4 = 1.6 cub. ft. Thus A x 3 = 1.6 cub.ft., where A is the sectional area of the pipe-line. phi/4 d 2 = A = 1.6/3 = 0.533, where d = diameter of pipe-line. d = root 4 x 0.533/phi = 0.825 ft. ( or 9.9 in.). Energy of wsater in jet = initial energy of water. The enerfgy of water to start with is simply h = 300, since P/w = 0 andv 2/g2 = 0. further utilises the unconverted head or potential energy of water, but the Pelton wheel makes use of this same energy after it has been converted into the more convenient form of kinetic energy. An hydraulic engine with a cylinder and piston could be used instead of either machine, but this would take the water at the lower end of the pipe-line where its velocity energy is practically nil and all the original head energy has been converted into the intermediate form of pressure energy. It could only do this, of course, if the nozzle were shut..this could be piped to a hydro-electric station in the manner shown in the bottom picture. Electrical energy derived from water-power can be turned into heat or light, as in domestic appliances, or4 it may be turned back into mechanical energy, as in an electric train. The picture facing shown it being employed to re-create potential energy by causing the ascent of a car on a rack railway.This force of 10 lb. is the resistence that has to be overcome in ordewr to drag the block up the slope. The work done by the applied fotce is therefore 10 x 100 = 1000 ft.lb. The other way of doing this is to say that 50 lb. has been raised vertically through a height of 100/5 = 20 ft. so that the work done = 50 x 20 = 1000 ft.-lb. What difference would it make if, in the last problem, the slope were not smooth-if the coefficient of friction between this and the iron block = 0.25? There is only one way to do the problem now, and that is the first way, allowing for the extra work done in overcoming friction by adding this the 1000 ft.-lb. that must still be done in opposition to gravity. The component of 50-lb. weight perpendicular to the slope now concerns us; it is root 50 2 - 10 2 = 49 lb. Frictional resistance to sliding = 0.25 x 49 = 12 1/4 lb. Work done in overcoming this = 12 1/4 x 100 = 1225 ft.-lb. Therefore the total work done in dragging the weight = 1000 + 1225 = 2225 ft.lb.