A short Account of Horizontal Water-Wheels. — 26: 
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The following Problems may sometimes be useful : 
Pros. ]. Given the angle between two cuts, to find the number 
of cuts. 
Rule. Find the angle in the table, and against it stands the 
number. 
Pros. 2. Given the number of cuts, to find the angle be- 
tween two. 
Rule. Find the number in the table, and against it stands 
the angle. 
Pros. 3. Given the angle between two cuts and the radius 
of the wheel, to find the width of a cut, 
Rule. Multiply the versed sine of the angle by the radius, 
and the product is the width of a cut. 
Pros. 4. Given the number of cuts and the radius of the 
wheel, to find the width of a cut. 
Rule. Find the versed sine (against the number) in the 
table, and multiply it by the radius for the width. 
Pros. 5. Given the angle between two cuts and the width 
of one, to find the radius of the wheel. 
Rule. Divide the width of the cut by the versed sine of the 
angle, and the quotient is the radius. 
Pros. 6. Given the number of cuts and the width of one, 
to find the radius of the wheel. 
Rule. Find the versed sine (against the number) in the table, 
by which divide the width, and the quotient is the radius. 
Pros. 7. Given (D) the depth of the fall, and (a) the dia- 
meter of the wheel, both in feet, or both in inches, to find the 
number of revolutions in a given time. 
Rule. Take sen, = 2% =n = number of revolutions 
in a second; then x X number of seconds in the given time 
gives the number of revolutions in that time. 
Example. Let D=45 feet and d=5 feet; then 
V/ 45 = 5230 x 5 = 2°562 revolutions in a second =”, and 
2°562 x 60= 153-72 revolutions in a minute. 
Pros. 8. Given (D) the depth of the fall, and (7) the num- 
ber of revolutions in a given time, to find the diameter of the 
wheel. : 
Rule. Find (n) the number of revolutions in a second: 
Then since Ny .*. = —— sd, 
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Example. Let D=30 feet, and the number of revolutions 
in 
