Kindly keen your queries on separate sheets of paper when corresponding about such matters as patents. subscription^ books, etc. This will greatly facilitate answering your questions. as in many cases they have to be referred to experts. The full name and address should be given on every sheet. No attention will be paid to unsigned queries. Full hints to correspondents are printed from time to time and will be mailed on request. (12555) J. F. S. asks: B is a brass cylinder, 2 inches diameter by 12 inches long. e is a plunger, weighing %, pound. D is a %-inch pipe connected at bottom of cylinder B3 leading to tank 4 inches above top of cylinder. Why won't plunger G lift water or force it rather into box, as the plunger is much heavier than the water it has to raise ? What size pipe will it require, as I cannot increase weight of plunger ? A. The true height of lift in your sketch is not 4 inches, but the total height from base of plunger where it rests on water to level of bend in pipe. A %-pound plunger in a cylinder of 3,14 inches area (equal to 2 inches diameter) will balance a head equal to % 3.14 pounds or 0.24 pound per square inch nearly, equivalent to 6.6 inches of water (since 12 inches head of water is 0.434 pound per square inch pressure). From your sketch we judge your plunger is about 6 inches long, and that the head of water is therefore about 6.6 inches at the height to which it is shown as rising, while you would have to add 'h pound more weight to the plunger to lift it to the tank (disregarding friction). You can reduce the diameter of cylinder and plunger to 1.6 inches nearly , and the head will then just balance the %-pound plunger. (12556) J. M. K. asks : 1. How much loss of power would result in transforming say 10 volts, 10 amperes, direct current into 20 volts, 5 amperes? Have you got a Supplement describing such a transformer, and the making of same ? . large loss would take place in changing 10 amperes at 10 volts direct curent to 20 volts. It could not be changed to 5 amperes at 20 volts. There is no transformer for a direct current excepting some form of an induction coil which will give an interrupted direct current. It would be cheaper and simpler to double the battery and so get your 20 volts and 5 by a suitable resistance. 2. Have you got a Supplement describing a small voltmeter? Also a small ammeter? A. You will find a voltmeter and ammeter described in our Supple-mknt 1215, price ten cents. They are of simple construction and can be made by an amateur. 3. Have heard that the Mississippi River flows up hill, being driven onward by the centrifugal force of the earth ? A. The expression that the Mississippi River flows up hill because of centrifugal force is not a correct expression. Water cannot flow up hill. A water level is the of still water anywhere on the earth. The ocean is level, but at the equator the water and the land surface also are thirteen miles farther from the center of the earth than at the poles. The Mississippi River at its source is above the level of its mouth, and the water flows down to its mouth from the source. Centrifugal force lifts the water level thirteen miles at the equator above its position at the poles. A ship does not sail up hill from New York to the equator. It goes over a water level, and neither up nor down. (12557) I. M. says : Kindly let me know whether the wheels of an automobile on the inside or outside of a curve leave the ground when going at great speed around said curve, anrl why? A. The inner wheels of a vehicle lea ve the ground first when it overturns in rounding a corner too rapidly. Centrifugal force acts upon the vehicle through its center of gravity, which is above the ground. The fulcrum is on the ground through the rim of the wheels. This forces the vehicle to turn over upon the outer wheels. The outer' rail of a track is raised to prevent the same effect in railroad trains. The outside of a race track is banked for the same purpose in bicycle races. The riders lean in very far in rounding the curves to prevent being thrown outward.