4. Divide the circumference of the wheel in feet, by the velocity of its floats in feet per second, and the quotient will be the number of seconds in which the wheel turns round.

5. By this last number of seconds divide 60; and the quotient will be the number of turns of the wheel in a minute.

6. Divide 90 (the number of revolutions which a millstone of 5 feet diameter ought to have in a minute) by the number of turns of the wheel in a minute, and the quotient will be the number of turns the millstone ought to have for one turn of the wheel.

7. Then, as the number of turns of the wheel in a minute, is to the number of turns of the millstone in a minute, so must the number of staves in the trundle be, to the number of cogs in the wheel, in the nearest whole numbers that can be found,

8. Multiply the number of revolutions performed by the wheel in a minute, by the number of revolutions made by the millstone for one of the wheel, and the product will be the number of revolutions performed by the millstone in a minute,

In this manner the following table has been calculated for a water wheel of fifteen feet in diameter, which is a good medium size, the millstone being five feet in diameter, and revolving 90 times in a minute.

To construct a mill by this table, find the height of the fall of water in the first column, and against that height, in the sixth column, you have the number of cogs in the wheel, and staves in the trundle, for causing the millstone to make about 90 revolutions in a minute, as near as possible, when the wheel moves with half the velocity of the water: and it appears by the last column, that the number of cogs in the wheel, and staves in the trundle, are extremely near the truth.

On the motion of bodies in fluids.

PROPOSITION CXCIV.

765. To determine the resistance of a fluid to any body moving in it, having a curved end, as a sphere or a cylinder with hemi-spherical end, &c.

Let BEAD (fig. 229.) be a section through the axis CA of the solid, and let the body move in the direction of that axis, or in the direction CA. To any point E of the curve, draw the tangent EG, meeting the axis produced in G; also draw the perpendicular ordinates EF, ef, indefinitely near to each other; and draw ae parallel to CG.

Put CF x, EF=y, BE=2, $sine<G to radius 1, and p=8.14159, &c. then 2py is the circumference of a circle whose radius is y≈EF, and 2py × Ee=2py ׿, is the fluxion of the surface described by Ee in revolving about the axis CA.

Now by art, 522, if the surface generated by Ee were a plane surface moving perpendicularly in a fluid, the resistance against it would be equal to the weight of a column of the fluid whose base is the surface resisted, and altitude the space through which a body must fall from rest, in vacuo, to acquire

the velocity of the plane: that is, ×2pyż would be the 4g

resistance against the surface if it were a plane perpendicular to the direction CA of its motion; but Ee is inclined to CG in an angle whose sine is s; therefore, (art. 515.) if n be the

specific gravity of the fluid, nxs3 X

v2

4g

xyz.is

2g

the whole resistance against the surface generated by Ee; or the fluxion of the resistance the body meets with, whatever may be its figure; and its fluent will be the resistance required,

766. Cor 1. When BEAD is a semicircle, or the form of the body is spherical; put the radius CA or CB=r, and we

EF CF

have y=√(r2—x2), s=EG CE and ży=EF×Ee=

generated by the arc BE; and when CF is CA, or x=r, it

Xa, is the resistance to the spherical surface

is the resistance which a cylinder meets with, when moving in the direction of its axis; now, since the diameters of the cylinder and globe are equal, and this latter expression, or

is double the former, or

pnd2v2
32g

it shows that the

16g resistance to the sphere is equal to half the direct resistance to a great circle of it, or to a cylinder of the same diameter; the same as is found by a different process in art. 523.

768. Cor. 3. Because pd3 is the magnitude of the globe, if N denote its specific gravity, its weight will be=}pd3N;

and, therefore (art. 251.) the retarding force F=

M_pnv3d2 Q32g

and this, by the theorems in art. 287. is

N

X

4gs

and sxd the space that would be

described by the globe, while its whole motion is generated or destroyed by a constant force, which is equal to the force of resistance, if no other force acted on the globe to continue its motion.

And if the densities of the globe and fluid are equal, the resisting force is such, as, acting constantly on the globe without any other force, would generate or destroy its motion

4d

in describing the space by that accelerating or retarding

force.

34

769.

PROPOSITION CXCV.

To determine the relations of velocity, space, and time of a globe moving in a fluid, in which it is projected with a given velocity.

Let a the velocity of projection, s the space described in the time t, and the velocity at that time; then (art. 768.) 3nd2

the accelerating force F

globe, n that of the fluid,

16g Nd

N being the density of the

and d the globe's diameter.

Now (case 4. art. 661.) the general equation vi=2gfi,

-Snv
8Nd

becomes vv=- Xs, or dividing by v2,

And the correct fluent of -=—bi, is

bi, putting b

3n 8Nd

log. a-log. v, or log.bs; or putting c=2.7182818, the

number whose hyp. log. is 1; then we have c, hence v=

=ac-bs

770. Cor. The velocity v, at any time t, being the cbs part of the first velocity, the velocity lost in any time will be the 1-c-bs

part, or the

cbs-1

cbs

part of the first velocity.

EXAMPLES.

Er. 1. If a globe be projected with any velocity in a medium of the same density with itself, and it describe a space equal