This rule is easily seen from the figure. FC G, represents part of the curved surface of the earth, B C the elevation of observer, AB E the Dip. Then ABEE BH, the angle between the water-line of object and the horizon, A B H, will, for such a short distance, be equal to ▲ BH C, the angle of elevation for which the constant is calculated.

=

These constants can also be used for the angle of elevation if you know the height of an object.

Example 1.-The height of the lantern at the South Foreland low lighthouse is given in the Admiralty list of lights as 275 feet above the sea; on passing it I observed the of elevation of the lantern above the water to be 2° 5', required my distance from it?

Constant for 275 × 13 ÷ 23 =

miles, the required distance.

155, and 2° 5' = 125′ ÷ 155 = 1.24

Example 2.-Coming round the North Foreland from the Thames, and not seeing the Elbow buoy which lays 2.4 miles from the lighthouse, and wishing to pass half-a-mile eastward of the buoy, or, say 3 miles from the lighthouse, I find the height of lantern at the lighthouse is 188 feet above the water, the constant for which will be 188 x 13 ÷ 23 = 106 constant, which 3 miles 35' to get on sextant, which will carry me 3 miles from the lighthouse, and half-a-mile eastward of the buoy and shoal.

=

This is a reverse application to the first example.

If a constant for each lighthouse were marked in the published list of lighthouses, their distances could be readily got on passing them at a moderate distance.

It is also handy to make out a table from the constant you use for your own height above the water in using the angle of depression. I also append a Table of Constants for heights up to 100 feet, which may be useful to some persons; if a number above 100 is required, add any two given numbers together that will make the required number, and the sum of their constants will give the required constant; or multiply the number of feet and its constant by any figure that will produce the required

number.

Yarmouth, October, 1875.

T. E.

A TABLE OF CONSTANTS FOR DIFFERENT HEIGHTS.

921

This table is constructed similar to Raper's Table 9, except that this is calculated to whole feet. The first number shows that 1 foot will subtend an angle of 1 minute at 0.565 of a nautical mile, and all the others are in like proportion.

INMAN LINE. QUICK PASSAGE.-The steamer City of Berlin made her last passage from Roche's Point to Sandy Hook in 7 days, 18 hours, and 2 minutes, mean time; and from Sandy Hook to Queenstown in 7 days, 15 hours, and 28 minutes, mean time. Her best day's run out was 381 miles, and her best day's run home was 388 miles. owners claim this as being the fastest run out and home ever made The across the Atlantic. The builders of the vessel are the Messrs. Caird, of Greenock.

T

DIRECT-ACTING SPRING SAFETY-VALVES.

HE reader in the Nautical Magazine for September, in going over the article on Spring Safety-Valves, will have no difficulty whatever in seeing from the formation of the valve and

its concentric chamber around the head, thereof, that the valve will lift well from its seat, and allow the steam to flow copiously from the boiler, by reason of the increased area on which the steam acts after it passes the seating orifice. But when he examines the "Table of Performances," and sees the rapidity with which the valve returns to its seat, he will then encounter the difficulty; and when I tell him that in order to obtain such a performance, the law of gravity must be reversed, that the force acting on the lower side of the valve at the instant when it closes is greater by two or three hundred weight than that which first raised it from its seat, he will probably stand aghast, and display the character of misbelieving" Thomas." But his unbelief will not now alter the fact accomplished, for thirteen hundred valves now on board ship (every time they rise from their seat to relieve the boiler) all cry out with one accord, "It is true!" and close again, faithful to the task assigned them.

ON THE BEHAVIOUR OF THE STEAM IN THE CONCENTRIC CHAMBER C. Let Fig. 3 be a valve applied to a boiler loaded to 100 lbs. per ", and

with which gauge No. 4 is in direct communication. Let gauge No. 3 be applied to the lower side of the chamber, No. 2 in the neck, and No. 1 on the surface of the valve. On the instant of No. 4 showing 100 lbs., the valve will rise from its seat, and the escaping steam will register on the gauges applied to the chamber C the respective pressures 25, 42, and 15 lbs.; gradually will all those pressures fall until, when about 12 lbs.,

the three gauges Nos. 1, 2, and 3 all become alike, and instantly the valve falls to its seat, and No. 4 gauge will be about 99 lbs. ; and the time occupied in making this change varies from 10 to 40 seconds, according to the size of valve, a 3" valve doing it in 10 and a 5′′ doing it in 40 seconds. But if a gauge be fitted to the stricture orifice, which is the orifice leading from the chamber C into the waste-pipe, and at right angles to the path of the escaping steam, then that gauge shows a line of no pressure, or an atmospheric line, at the instant when the valve closes; and if the stem of the pipe on No. 3 gauge be carried through the chamber C, and the end terminating in the stricture orifice, by applying a vacuum gauge 4 lbs. of a vacuum is to be found at the instant when the valve closes.

ON THE MAGNITUDE OF THE FORCE ACTING ON THE LOWER SIDE OF THE VALVE AT THE INSTANTS OF RISING AND FALLING.

Take a 6" valve, area 28-27□", and let the diameter of the stricture orifice be 81", area 50·26", then 50-26 28-27 = 20□", nearly equal area of concentric head on valve. Let the absolute pressure in the boiler be 114.7 lbs. at the instant the valve rises, and 113.7 lbs. at the instant of closing, then the total force acting on the lower surface of the valve tending to raise it from its seat is equal to 28.27 × 114·7: 3,242 lbs. ; and at the instant the valve closes, we have a force acting on the lower surface equal to 28.27 × 113.7 3,214 the absolute pressure acting on the concentric area of head of valve, and shown by gauges Nos. 1 and 3 at the instant of closing, and is equal to 26·7 × 20 = 534 lbs., and 534 + 3,214 3,748 lbs., equal to the total force acting on the lower surface of the valve at the instant of closing, and 3,748 – 3,242 lbs., equal the excess of force acting on the lower surface of the valve at the instant of closing, and greater than that which first rose it from its seat.

506

This is the result of practical observation and practice, and the passive observer, or practical-minded man, could go no further into this subject. Hither can he come, but no further; and not until the philosopher came to his aid could he ever discover the cause of this unique phenomenon.

At the instant when the gauges 1, 2, and 3, all become of one pressure -viz., 12 lbs., the gauge which is applied at the stricture orifice will show a line of no pressure, or atmospheric pressure, and the valve exists in the state of the balance, notwithstanding there is an excess of force on the lower surface of the valve equal to 506 lbs. The fulcrum of the force is in the centre of the current of the issuing steam between the seating orifice and the stricture orifice, and at the instant of the valve closing, that line is neutral, and the valve being up from its seat, and the spring compressed beyond its load, it has a force at its command equal

to its extra compression for the purpose of closing the valve, which it very quickly puts into action, and the valve falls to its seat all at

once.

STEAM, CAN BE MADE TO PASS THROUGH AN ORIFICE AT A VELOCITY GREATER THAN THAT DUE TO THE INTERNAL PRESSURE.

Fig. 9. Let O be any orifice for the escape of steam, let R t, R' t', be

the boundary walls of the orifice, and let there be three molecules, T, D, H—"Tom," "Dick," and "Harry "-attempting to escape from prison at the same time. Now, Tom is next the orifice through which they are about to escape, and he is unbalanced on the side next O by a space equal to 120° of his periphery; owing to his unbalanced condition he would roll out on the side O, but he would take his own time to it, and "Dick" and "Harry," are impatient to get out, and they are each equally as strong as Tom, so they both unite with their full force to push Tom out; but they cannot both get direct behind him to push him straight out, so they apply their efforts at an angle of 30° each from the line of escape; this reduces their force to 866 of the full magnitude, or the cosine of the angle of application F, P, P", but ⚫866 × 2 = 1·732. Now, let the velocity of Tom, due to the internal pressure, be represented by radius or unity, then the velocity with which he will be sent out into the atmosphere will be represented by 1.732 F P. Or if the velocity of steam, due to pressure, be 1,000 feet per second, then by the conditions herein set forth the velocity at the escaping orifice will be 1,000 × 1.782 = 1,732 feet per second.

ON THE NEUTRAL LINE OF FORCE IN THE CENTRE OF THE ISSUING

CURRENT OF STEAM.

Let T, D, H be again three molecules of steam under pressure, but not allowed to escape, the fulcrum of the forces is in the centre of the intersticial space between the molecules; let the orifice O be suddenly opened, T will escape, and D and H will follow, but they cannot go out in a straight line, because they meet with partial resistance from the walls of