Q. About how much is the average engine friction? A. According to the American Locomotive Co., 22.2 lbs. per ton, or 1.11 per cent of weight on drivers. Q. How much is the head air resistance? A. It is equal to the cross sectional area (say 120 sq. ft.) times 0.002 of the speed in miles per hour. Q. How much is the resistance due to the weight on trucks, trailers, and tenders? A. The same per ton as that due to the cars. Q. How much engine and tender resistance does the grade make? A. 20 lbs. a ton for each per cent of grade. Q. With what does the curve resistance to engine and tender vary, and how much is it? A. With the engine wheel base; for each ton it is 0.4 plus the product of the degree of the curve by the following constants: Wheel base, ft... 5 6 7 8 9 12 13 15 16 20 Constant 0.380 .415 .460 .485 .520 .625 .660 .730 .765 .905 Q. What is the resistance of freight cars? A. The Penna. R. R. figured it as follows: Tons, per car. 10 20 25 30 40 50 60 70 72 Resistance, lbs. per ton. 13.10 7.84 6.62 5.78 4.66 3.94 3.44 3.06 3. TRAIN SPEED Q. How can the speed be calculated where the distance between points, and the time consumed, are known? A. Multiply the miles by 60 and divide by the minutes. Thus 5 miles in 6 minutes 5 X 606 = 50 miles an hour. Q. What is the distance between telegraph poles in the U. S. and Canada? A. 55 yards or 32 per mile, so that the number of poles passed in 19 seconds gives nearly the speed in miles per hour. Q. How many rail lengths to a mile? A. Usually 5,280 30 176. Q. What is the quick method of approximating the speed of the train on which you ride? = A. If the rails are 30 feet long, court the number you pass in 20 seconds and you will come very near the speed in miles per hour, as 5,280 30 176, and 3,600 20 ÷ ÷ 180, which is not so far out. A trifle nearer would be to count the number of rails in 41 seconds and multiply by two; thus 3,600 41 87.8; which is practically the half of 176. Q. How else than as shown on page 457 can you determine the speed of a train in miles per hour? A. By multiplying the driver circumference in inches by the turns per minute, and either multiplying by 0.00947 or dividing by 1,056. Thus if the drivers are 194 inches in circumference and the driver makes 150 turns a minute, the speed will be 194 X 150 X 0.00947 = 27.56 miles an hour; or 194 X 150 1,056 gives the same re sult. Q. How do you get the circumference of the drivers? A. Either by using a tape measure or by multiplying the measured diameter by 3.1416. Q. Is there any other way of estimating train speed? A. Yes; half the number of 30-foot rails passed in 41 seconds, or of 33-foot rails in 45 seconds, will give a close approximation to speed in miles an hour. Thus, sixty 30-foot rails in 41 seconds makes 60 X 30 X 60 X 60 divided by 41 X 5,280 29.9 miles, practically half of 60; and sixty 33-foot rails in 45 seconds makes 60 X 33 × 60 × 60 divided by 45 × 5,280 = 30 miles (exactly half of 60). Q. Give a table showing how many turns a minute are necessary for various train speeds and driver diameters. A. As under: Q. How would you get such figures for other wheel diameters and train speeds? A. Multiply the miles per hour by 5,280; divide the product by 60 times the wheel circumference in feet (equal to the circumference in inches multiplied by 0.2618). Q. Can you give a table showing the number of wheel turns that are necessary for various wheel diameters in making a mile run, not allowing for sliding? A. Here is such a table, the figures rounded off to the nearest whole number: Q. Why do you give 100 when there are no 100-inch drivers? A. To facilitate finding the turns needed for wheels of other diameters than those in the table. Q. How do you find the needed driver speed for various wheel diameters and train speeds? A. To get the turns per mile (5,280 feet) divide 1,680 by the wheel diameter in feet. To get turns per minute multiply the miles per hour by 28 and divide the product by wheel diameter in feet. Thus an engine with five-foot drivers, at 40 miles an hour, would make 1,680 ÷ 5 = 336 turns a mile or 40 X 28÷ 5 224 turns a minute. Q. Can you make a table showing piston speed in feet |