An Easily Constructed High Tension Insulator
��Popular Science Monthly 959 '
Finding the Combined Value in \ Parallel Resistances
��THE drawing below shows a high tension insulator that is easily made, has good strength and insulation, and is businesslike in appearance. It can be made for about five cents. Two cleats are used for average power in ama- teur stations bolt&
NUT S
��CLEATS
���METAL STRIP
An easily constructed insulator for use on high tension wires
��while three or more can be used for others of greater potential. The cleats are joined with y%-m. iron or brass strips fastened with small bolts as shown in the sketch. For good appearance all the metal parts should be enameled black, . while the porcelain can be covered with thick shellac, giving it the appearance of brown glaze. — John B. Rakoski.
��Heating Hard Sheet Rubber to Facilitate Cutting Disks
THE amateur experimenter and worker on electrical apparatus at some time or other desires to cut hard sheet rubber into round disks for making Wimshurst machines or other devices. Hard rubber, as received from the stock house, is very difficult to cut, unless you have all the facilities for doing such work.
The writer has found a very easy way to cut hard sheet rubber up to 3/16 in. thick. The method is as follows: Mark on the rubber with a scriber or other sharp instrument an outline of the piece to be cut. Then plunge the sheet into hot water; take it out and cut on the outline with a pair of scissors. The rubber will become soft like leather and cut easily. As it becomes cooler, it will cut harder. If any more cutting is to be done, plunge it into hot water again and continue until the cutting is done. Then put a plate of glass on a newspaper on a table, dip the rubber in the hot water again, place on the glass; then put another sheet of glass over the rubber with a weight on it. When it is cooling the piece will straighten out. — W. S. Standiford.
��WHERE more than two resistances ] are connected in parallel it usually results in rather complicated calculations in order to determine the combined re- sistance value. It has been found that an easy way to determine the equivalent of resistances in parallel is by the use of a simple diagram, as shown. •
As an illustration, the four resistance values, 225, 450, 900 and 1800 ohms were taken. The procedure is as follows: Decide on some convenient scale to be used. In this instance, each 1/10 in. equals 50 ohms. Lay off a perpendicular line X- Y to represent any one of the resistances, the one of the highest value, which in this case is 1800 ohms, probably being the most convenient. Then construct the horizontal line Y-Z of any convenient length- and erect the perpendicular A-Z equal in height to the next highest re- sistance. On this line lay off the remaining resistances according to the scale selected. For convenience we xisoo will start with the two
highest values, although the same results fwould be accomplished by us- ing the 1 800-ohm resis- tance and any one of the other three shown on line A-Z. Draw
���A convenient scale for finding the com- bined value of resistances in parallel
lines X-Z and A-Y. At their point of intersection, B, erect the perpendicular B-C whose value represents the equivalent of the 900 and 1 800-ohm resistances in parallel. Combine this equivalent with the 450-ohm resistance by drawing line H-C. At the point of intersection D of this line with the line B-Z, which happens to fall on the line X-Z, erect the perpen-
�� �
