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blocks of many substances on account of their shape or physical properties, and many other bodies are too small or too valuable. We cannot, then, compare their densities in this way, but we may take some substance as a standard, and compare the weights of all others with this.

Now any substance might be chosen for this purpose, the main requisites being that it shall be easily procurable in a state of purity, and easy of manipulation. Water has been chosen as this standard, and is found to answer well. When, therefore, we speak of the specific gravity of any body, we mean this-the proportion which exists between its weight and that of an equal bulk of distilled water at a temperature of 60°.

The reason why we thus fix on a certain temperature is that water expands by heat, and therefore a cubic inch of hot water weighs less than an equal bulk of cold. The temperature of 60 is chosen merely as a matter of convenience, that being about the average, and therefore involving less trouble. When, then, we say the specific gravity of mercury is 136, we mean that any amount of mercury weighs 13'6 times as much as an equal bulk of distilled water at 60°. Now, as we have seen, the weight of a cubic inch of distilled water is 252.5 grains; a cubic inch of mercury therefore weighs 2525 grains X 13'6, or 3,434 grains, which is nearly 8 oz. We can in this way, if we know the specific gravity of a body, tell the weight of any bulk of it. Questions like the following frequently occur, and can thus be solved :-What is the weight of a block of coal 3 feet x 5 x 4, the specific gravity of coal being 1.270 ?

Since the specific gravity of the coal is 1.270, the weight of a cubic foot is 1.270 times that of an equal bulk of water. But a cubic foot of water weighs about 1,000 oz.; a cubic foot of coal must then weigh 1,270 ounces. Now the total bulk of the coal is 3 x 5 X 460 cubic feet. Its weight, therefore, is 60 x 1,270 ounces=76,200 ounces, or 42 cwt. 2 qr. 21 lb. Again, strong oil of vitriol has a specific gravity of 1850; how much will 6 lb. measure? A fluid ounce of water, it must be remembered, weighs one ounce avoirdupois; 6 lb. of water, then, would measure 96 oz.: but since oil of vitriol is heavier than it, in the proportion of 1,850 to 1,000, it will measure proportionately less. Hence the following proportion will give us the bulk:

As 1,850: 1,000 :: 96: the required volume..

On working this out, we shall find that the vitriol will measure 51-89 oz., or nearly 3 ordinary pints.

We see thus the importance of knowing the specific gravity of any substance, and there are several modes of ascertaining it, one or other of which is more applicable according to the circumstances of the case. If, however, we bear in mind exactly what it is we wish to know, we shall find little difficulty in remembering which way to proceed.

We will consider, first, how to proceed in the case of a liquid. Procure a thin glass flask (Fig. 13, A) provided with an accurately fitting stopper. Instead, however, of this being solid, let it be drawn out, as shown at B, into a long tubular neck, so that when it is put in its place any excess of liquid may escape through it. The flask is best made of such a size as to hold 1,000 grains up to the mark o in the neck. Now procure a small piece of metal, and file or grind it till it exactly balances the flask and stopper when empty. If the flask, filled with distilled water to the level o, be put in one pan of a balance and the counterpoise or weight in the other, we must add just 1,000 grains to balance the water. Empty this out, and fill it with the liquid whose specific gravity we want to know-say, for instance, the strongest alcohol-and weigh again; we shall now find that only 792 grains are required to balance it. The weight, then, of any volume of alcohol is to that of an equal bulk of water in the proportion of 792 to 1,000; or, in other words, the specific gravity of the alcohol is 792. The reason why we chose a flask containing 1,000 grains is now clear, for all trouble in calculation is thus avoided. We have only to take the weight of the liquid in grains, and point off three figures as decimals, and we have the specific gravity.

Thus, if we fill the bottle with sea water, we shall find it will weigh 1,028 grains; the specific gravity is therefore 1.028. Sometimes, however, it is difficult or costly to procure a #afficient quantity of the liquid to fill such a large flask. We then use a much smaller one, usually made out of a glass tube,

and, as before, we first find the weight of water required to fill it to a certain mark, and then the weight of the liquid we are operating upon.

The details of an actual experiment will make this clearer. A sample of nitric acid was taken, of which it was desired to ascertain the specific gravity. The small bottle was first put in the scales and found to weigh 80 grains. On being filled with the acid it weighed 159 grains. The acid was next emptied out, the bottle rinsed, and filled to the same height with water, the weight being then 136 grains.

Fig. 13.

Now, since the bottle weighed 80 grains, we subtract this amount from its weight when filled with the different liquids, and thus see that the water in the bottle weighed 56 grains, while the weight of the same bulk of acid was 79 grains. We have, then, the following equation by which we can determine the specific gravity of the liquid:

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ANSWERS TO EXAMPLES IN LESSON II. (Vol. II, page 398).

1. Since the areas of the pistons are to one another in the proportion of the squares of their diameters, the larger has 144 times the area of the smaller. There is also a gain of 6 by the lever. Thus the advantage gained is 144 x 6 or 864. If we divide 20 tons by this, we find the required pressure is 51-85 pounds.

2. The area of the sides is together 144 feet, and the mean depth 2 feet. The total pressure on them is thus equal to the weight of 9 tons and 90 pounds. The pressure on the bottom is 6x10 × 4×1,000 144×24 or 324 cubic feet of water. This is nearly 324,000 ounces, or ounces, or 7 tons 10 cwt. 2 qrs. 19 lb.

3. Rather more than 19; tous.

4. The area of the large piston is 38 square inches. The required pressure is therefore nearly 40 pounds.

5. Just over 15 tons.

6. The additional pressure will be that of a column of water having an area of 6 square inches, and a height of 3 feet. This will be of The total pressure is a cubic foot, and therefore weigh 7 lb. 13 oz. therefore 10 lb. 13 oz.

LESSONS IN GREEK.-XIV. REVIEW OF THE THREE DECLENSIONS. WITH the nouns of the first and second declension, the student, if he has thoroughly mastered the foregoing lessons, will find no difficulty in any attempt he may make to construe classical Greek. It is somewhat different with nouns of the third declension, the discovery of the nominative of which is necessary in order to consult a Greek lexicon with case and effect. I therefore subjoin the following, which will enable him from the genitive case to find the nominative; in which form substantives and adjectives appear in dictionaries. I give the genitive, because the genitive is, as it were, the key to the remaining oblique cases. Thus, if you meet with avopa, you know the genitive must have two of these letters, namely, 8p; if you meet with Xeuwves, you know the genitive will have the letters χειμων ; if you meet with μελανες, you know the genitive will have the letters peλav. Now, from the genitive you may get

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11. Прoσтide, adds, from рoσтionμi, I add; emiotnun, ins, n, understanding.

12. Boνkepw, having the horns of an ox, from Bouкepws, -w, and that from Sous and kepa; lous, Io, from Iw, -oûs; #λavai, wanderings, from πλανη, ης, η.

14. Aλnows, truly; ws aλnows, very truly.

15. 'Holoty, sweetest, the superlative degree of hous, sweet; porpopor, pleasant, from #poσqupos, -ov, suitable to (pos and фера).

16. Apiary, the best, that is, noble, from apioros, a superlative of ayados.

17. Ξίφος, τους, το, a sword; τιτρώσκει, wounds, from τιτρώσκω, I wound.

18. MeyiσTov, the greatest, superlative from μeyas, great. 20. Tuparvis, -idos, n, usurped power, tyranny; adikias, of injustice (a privative, and dikη, right, justice).

21. Δείλος, -η, -ov, cowardly, ὁ δειλος, the coward; προδοτης, -,, a betrayer, traitor.

22. Elkoves, images; elкwv, -ovos, ó, an image.

23. Nouades, the nomads, or wandering tribes, from vouas, -os, and that from veuw already explained; apilμovσiv, they Lumber, from apilμew, I number, our arithmetic.

24. Exovoar, having, present participle from exw, I have; it agrees with γαστερα.

25. Ήφαιστος, Vulcan; χωλος, -η, -ov, lame.

26. Mndela, -as, n, Medea; viжоßλeжоνσα, scowling at, from vro, under, and Bλεлw, I look. (Compare the Latin suspicari: sub, specio.)

27. Ήθους, of character, from το ηθος ; βασανος, -ου, ή, a touchstone, test.

28. Opis, odews, d, a serpent; ios, -ov, a dart, sting.

29. Пapvaσoos, Parnassus, a mountain of Phocis, on which was Delphi; ovoKLOS, -ov, overhung with clouds, from ovv, with, and σkia, a shade.

30. Exionμos, -ov, distinguished, remarkable, from eπ, on (here an intensive), and σnua, a sign, whence our semaphore, that is, a telegraph ; Ἑλίκων, Helicon; Κιθαιρών, Citharon; καλουμενον, called, named, participle agreeing with To, that is opos; éTEPOS, ov, other, the other.

33. Ανάχαρσις, Anacharsis; είπε, said; ήδονης depends on BOTOUS: μEON, -ns, ǹ, intoxication; andia (from a, not, and nous, accet, disgust.

34. EURλela, -as, n, glory, distinction.

35. Decavos, -ov, d, Oceйnus, Ocean considered as a divinity; Theus, -os,, Tethys, a sea-goddess.

36. Σιτεομαι, I feed on ; δροσος, -ου, ή, dew,

37. KAearons, Cleanthes; eon, said; araidevтos, -ov, untaught, neducated; uopon, -ns, ǹ, form; diapepw, I differ.

35. Ονειδίζω, Ι reproach, Anacharsis being reproached; Σκύθης, a Scythian.

days, but by nights. 24. It is hard to speak to the stomach, as it has no ears. 25. Vulcan was lame in his feet. 26. Medea is painted as 27. Time is the test of men's scowling fiercely at her children. character. 28. Serpents have a poison in their teeth. 29. Parnassus is a great and shady mountain. 30. In Boeotia are two remarkable mountains, the one called Helicon, and the other Citharon. 31. The Nile has all kinds of fishes. 32. Honour your parents. 33. Anacharsis said that the vine bore three branches; one of pleasure, one of intoxication, and the third of disgust. 34. Toil is the parent of glory. 35. Inachus was the son of Oceanus and Tethys. 36. Grasshoppers feed on dew. 37. Cleanthes used to say that uneducated men differed in form only from wild beasts. 38. Anacharsis being reproached because he was a Scythian, said he was so in race but not in character. 39. In hell the bad are punished, (whether) kings, slaves, satraps, poor, rich, or beggars. 40. The daughters of Phorcus were old women even from their birth. 41. Zenon used to say that it was right to adorn

cities, not with monuments, but with the virtues of the inhabitants.

KEY TO EXERCISES IN LESSONS IN GREEK.—XIII. (Vol. II., page 390.)

EXERCISE 47.-GREEK-ENGLISH.

9.

1. Women rejoice in ornament. 2. The Greeks worship Zeus, and Poseidon, and Apollo, and other gods. 3. Modesty becomes women. 4. The dogs guard the house. 5. The pilot directs the ship. 6. The droppings of water make the rock hollow. 7. It is a woman's duty to watch her home. 8. It is the part of a good wife to keep house. The dice of Jove always throw luckily. 10. Dogs always afford men aid and pleasure. 11. The evidence of witnesses is often trustless. 12. Carry, my child, the key of the chest. 13. O Zeus, receive the prayers of the unfortunate man. 14. Castor and Pollux were the saviours of ships. 15. Silence brings adornment to woman. 16. The Ethiopians have dark hair. 17. O woman, preserve your house. 18. We comb our hair with a comb. 19. Eăcus keeps the keys of Hades. EXERCISE 48.-ENGLISH-GREEK.

1. Κοσμος πρέπει την γυναίκα. 2. Έργον εστι γυναικών φυλάττειν την οικίαν. 3. Depovi Kess The Oiklas. 4. Κλείδες της οικίας φέρονται τῇ μητέρα. 5. Τοις Αθηναίοις ησαν πολλαι νηες, 6. Διϊ ησαν πολλοί vaol. 7. Οἱ ιχθυες ανακύπτουσιν εκ του ύδατος, 8. Ο κυβερνητης ιθύνει την ναῦν. 9. Η ναυς εθύνεται ύπο του κυβερνητου. 10. Σεβεις Δια και Απόλλωνα.

EXERCISE 49.-GREEK-ENGLISH.

1. To drink much wine is an evil. 2. Kings have large revenues. 3. In Egypt is abundance of corn. 4. The sea is great. 5. Croesus had great wealth. 6. From a slight joy often arises great anguish. 7. To gentle words we yield with pleasure. 8. The great gifts of fortune bring terror. 9. The tempers of many men are gentle. 10. Toil is a great aid to virtue. 11. Children love gentle fathers and gentle mothers. 12. Keep up an acquaintance with gentle-hearted men. The women are gentle. 14. The majority of mankind call Alexander, King of Macedonia, Great.

EXERCISE 50.-ENGLISH-GREEK.

13.

1. Απέχου πολλου του οινον. 2. Οἱ κακοι χαίρουσι πολλῷ τῷ κοινῷ. 3. 4. Τοις βασιλευσι εισι μεγαλα Πολυς ὁ οινος βλαπτει τους ανθρώπους. προσοδοι. 5. Η πρόσοδος των βασιλέων εστι μεγάλη. 6. Αίγυπτος έχει πολυν σιτον. 7. Πολλοίς εστι πολυς πλουτος, ολιγος δε νοῦς. 8. Ορεγεσθε πραέων εθεών. 9. Τα εθη των γυναικών εστι πραέα. 10. Καλλος εστι προστεθεσι. 11.

39. Kolace, I punish; ev qdov, doμg is understood, in the abode of Hades, in hell; oarражηя, -ον, d, а satrap or governor ef = province ; πένης, -ητος, poor, πτωχος, -η, -or, begging, οἱ Αλεξανδρος, ο των Μακεδόνων Βασιλεύς, πολλακις μέγας προσαγορεύεται.

TYOL, beggars.

40. гpaia, n, old, an old woman, grey-haired.

41. Aer, that it was necessary, proper; avalημa, -тos, Tо, an chering, public monument, from ava, up, and Tienui, I place; TEP OLKOUTTOV, of their inhabitants, from oiкew, I inhabit (compare ouros and oikia).

KEY TO THE RECAPITULATORY EXERCISES FROM THE GREEK CLASSICS.

1. One swallow does not make a spring. 2. Time brings all things to light. 3. Atreus and Thyestes were the sons of Pelops. 4. Many things happen to men contrary to expectation. 5. Women's ornament sa (kind) disposition, not jewels. 6. Grasshoppers are said to be melodious. 7. Ants and bees have a laborious life. 8. Thief knows thief, and wolf (knows) wolf. 9. It is the use and not the possession ef books that is the means of education. 10. Nature without instruction is a blind thing, and instruction with nature is a defective thing. 11 Time brings knowledge to old age. 12. Many were the wanderings of the cow-borned Io. 13. Man saves man, and city city. 14. Very truly was Epaminondas a hero amongst heroes. 15. An old man has the sweetest tongue for an old man, a child for a child, and a woman

has a tongue suitable for woman. 16. All the children of the noblest Persians are educated at the king's court. 17. The sword wounds the body, but speech wounds the mind. 18. Intellect is the greatest good. 15. Laws are a city's soul. 20. Tyranny is the mother of injustice. The coward is the betrayer of his country. 22. Good men are the Exenesses of deity. 23. The nomad Lybians reckon (time) not by

EXERCISE 51.-GREEK-ENGLISH.

1. Speech is a mirror of the mind. 2. Men have intellect as a master. 3. Cherish a well-disposed friend. 4. Good friends have a faithful mind. 5. The voyage is uncertain to sailors. 6. Lead a life with discretion. 7. The mob has no discretion. 8. Do not quarrel with people. 9. The good are well disposed to the good. 10. Seck for good friends. 11. The bones of Orestes were in Tegea, 12. The female servants carry bread in baskets. 13. The gods give both the fair and foul voyage to sailors. 14. The intellect is the soul's curb. 15. Often the tempers of men reveal their abilities. 16. The speech of truth is simple. 17. A kind word lessens grief. 18. The cup is silver. 19. Death is called a brazen sleep.

EXERCISE 52.-ENGLISH-GREEK.

1. Ὁ νοῦς ἐστι διδάσκαλος ανθρώπου. 2. Ο εννοος φίλος θεραπεύεται. 3. Οἱ ευνοοι φιλοι θεραπεύονται. 4. Τοις εννοεις εισι πολλοί φίλοι. 5. Απέχου του ανίου. 6. Ορεγου των ευνοών φίλων. 7. Κομιζε τον αρτον εν τοις κάνοις. 8. Φεύγετε τους ανθούς νεανίας. 9. Οἱ νεανίαι αν οι φεύγονται. 10. Το κυπελλον εστι χρυσοῦν. 11. Τα κυπελλα αγγυρέα έστι καλα. 12. Βιον αγε τῷ νῷ. 13. Μη έριζε συν τοις ανθοις.

GEOMETRICAL PERSPECTIVE.-V. WE have said in a previous lesson that if we are able to determine the perspective position of one point, we can of more; and should these points be considered as the extremities of straight lines, we can, by drawing lines to unite them, represent the lines themselves. We repeat this statement for the

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purpose of placing before our pupils one or two cases to show how this principle, when put into practice, will simplify the work in many questions which may appear to be difficult. Let the pupil turn to Prob. I., Fig. 7 (Vol. II., p. 225)—of course we presume that this problem has been learnt, and, for the sake of practice, lines have been placed at other angles and worked out alsobecause this problem contains the key to the whole theory of ground-plan perspective. Now, if the pupil consider for a moment the principle exemplified in that problem of putting a hine only into perspective, he will see that it is nothing more than finding the positions of c and B. For as a straight line extends between the two points in the plan, therefore, when the perspective of these two points is determined by drawing a line between them, we shall have the perspective of the line. In order that we may be fully prepared to go further into the subject, we will take a point only, independent of any line, and place its plan anywhere beyond the P P, Fig. 27. Let A be the point. From A draw a line at any angle with the PP, and meeting it in a. Find the vanishing point of a A, bring down the point of contact a perpendicularly, to b, and join b v P; a visual ray from A cutting this line will give the perspective position of A at c. We will now make an application of this principle, and show how, with only one vanishing point, we may work out a problem which in plan is composed of a great number of lines, and none of them parallel with each other. We will propose a very simple case (Fig. 28), composed of three lines ab, be, and ed. Draw a line a e from a to the PP, and from the extremities of each line draw others parallel with a e; find the VP for one of these lines, which will be the V P also for the rest, because parallel retiring lines have the same V P; bring down the points of contact e, f, g, h, to B P, the base

of the picture,

VPL

9

40°

and then join them with the v P. Upon these last lines respectively will be the perspective positions of the points or extremities of the lines abed found by the visual rays as in Fig. 27, viz., a b'c' d'; unite these points by straight lines corresponding with the plan, and then will be produced the perspective view of that represented by the plan. The papil must practise this by giving himself a similar combination of lines, and of greater number. We will add another position of line to this figure. A line, a b, is drawn across the arrangement (Fig. 29). There is no necessity to repeat that which has been already shown in Fig. 28, so we will proceed to explain the method of representing the line a b. It will be understood that the perspective of c d is ed' and of e f, ef. Now the given line ab crosses these lines in the plan through m and n; it will be necessary, then, to find the projection of these points m and n, which will be done if visual rays are brought down from these points to cut the perspective lines dino, and e' f' in p; then, to complete the problem through these points o and p, the line a' b' must be drawn to the extent each way as determined by visual rays from the extremities a and b. These figures may in themselves have no particular meaning, we merely employ them to illustrate a principle, because the perspective of any other object of an intelligible character may be worked out by this method.

Fig. 30 is the plan of a hexagon (see Lessons in Geometry XIX., Problem LI., Vol. II., page 192), having one of its edges at an angle of 20° with the P P. We will only give the figure, and recommend the pupil to work it out; afterwards he should make other regular or irregular polygons for practice also. Remember that a regular polygon is one that has all its sides and

all its angles equal; an irregular polygon has unequal sides and unequal angles.

PROBLEM XII. (FIG. 31).—To draw the perspective of a circle. Diameter of the circle, 4 feet; height of eye, 5 feet; distance of the eye from the P P, 10 feet, and opposite the centre of the circle. Scale, inch to the foot. Draw the BP and also the H L 5 feet above it; anywhere (at a) draw the semicircle b c d, with a radius of 2 feet; from a erect a perpendicular to B P, and from the point where this line cuts the H L in P s describe the semicircle D E 1, E, DE 2, with a radius equal to the distance of the eye from the picture plane. About the semicircle deb describe the rectangle defb; draw a e, af, and through the points in the semicircle cut by the lines a e and a f, draw the lines giand hk parallel to a c or d e. From the points d, k, a, i, b, draw lines to the P S. Draw a line from d to D E 2, and also one from b to D E 1, and call them distance lines; notice where these last lines cut b P S and d P S, between these intersections the line mn must be drawn. We have now a square in perspective, in which the circle is to be drawn. Through the intersection of the distance lines draw o p parallel to m n. With the hand draw through the points t, u, p, q, a, r, o, s, to t again, the perspec

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tive circle required. This latter part of the process-that is, drawing the circle-will be the most troublesome; as it does not admit of the use of compasses, it must be done by the hand only, and will require much practice to do it neatly.

PROBLEM XIII. (FIG. 32) is the first example of elevations. A row of

7 rods, perpendicular to the ground, 6 feet high, and 1 foot apart; the plan 55° with the picture plane. Their distance apart must

VP.2

be ar

ranged in the plan 1, 2, 3, 4, 5, 6, 7; their heights must be set off on the line of contact. bc; therefore a line drawn from c to the V P will

represent their retiring position on the ground, and a line from b to the V P will represent the perspective of their heights; their widths apart will be found by the visual rays drawn from 1, 2, 3, etc., on the plan. PROBLEM XIV. (FIG. 33).—A cube 4 feet edge has one of its faces at an angle of 40° with the PP; its nearest angle touches the PP. Draw the plan of the cube a b c d, making de at an angle of 40° with PP. Find the vanishing points of the sides a d and d c, viz., V P 1 and V P 2. Proceed as in Prob. VII., Fig. 19, for the extent of the base. For the elevation, as the cube touches the P P at d, therefore a perpendicular line from d brings down the point of contact to e, making the line de the line of contact. As all heights are measured upon the line of contact, therefore the height of the cube must be represented by e f equal to d c (as all the edges of a cube are equal). Draw lines from ƒ and e to each vanishing point; then the visual rays from a and c will determine the outer and perpendicular edges of the cube g h and i k. Again, as parallel retiring lines have the same vanishing point (and opposite sides of a cube are parallel), draw a line from i to v P 1, and another from g to v P 2, where these lines intersect at m will complete the upper horizontal face of the cube, viz., gf mi. Make all these lines dark as in the diagram.

To find the perspective of the centres of any of the faces of the cube, it will be only necessary to draw the diagonals on the faces of the perspective projection-viz., from f to k and i to e, or from f to b and g to e. If a line is required to be drawn across the centre of a face, in a horizontal position, the above central point must first be found, and then the line must be drawn through this point to the VP of the face respectively.

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