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Velocity, whose ordinates vertical lines on a graph used to measure values B γ, B γ, and A O will give the particular velocities, and the S-shaped line of time will extend beyond N. The same result will occur if we assume the body (moved by the spring) is proportionately heavy, and the powers of the spring remain the same as in the first example.
And assuming the power of the spring is the same as at first, but bent only to point B 2 and released from there: B 2 E A is the triangle of its powers, the ordinates of the circle B g L are the lines of its velocity, and the ordinates of the S-shaped line B i F are the lines of time.
Having thus shown you how the velocity of a spring may be computed, it will be easy to calculate to what distance it will be able to shoot or throw any body that is moved by it. This must be done by comparing the velocity of the ascent of a body thrown with the velocity of the descent of gravity, while also making allowances for the resistance and impediment of the medium the substance, such as air or water, through which an object moves through which it passes.
For instance, suppose a bow or spring is fixed at 16 feet above a horizontal floor, which is nearly the distance that a heavy body will descend perpendicularly from rest in one second of time. If a spring delivers the body in a horizontal line with a velocity that moves it 16 feet in one second, then it shall fall at a point 16 feet away from the perpendicular point on the floor over which it was released. By its motion, it shall describe a parabola in the air or space through which it passes.
If the spring is bent to twice the former tension, so as to deliver the body with double the velocity in a horizontal line—that is, with a velocity that moves 32 feet in a second—then the body shall touch the floor at a point very near 32 feet from the aforementioned perpendicular point. The line of motion of the body so shot shall move in a parabola, or a line very near it. I say "very near it" because of the