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Plate I Fig. 1.
20. While a body moves with uniform motion Motion at a constant speed. along a curve AM, at each individual point M it will press normally Perpendicularly to the direction of the curve. against the curve with a force that is to the body’s force of gravity as the height due to its velocity In 18th-century physics, this is the vertical height from which a body would have to fall to reach its current speed ($v = u^2 / 2g$). is to half the radius of osculation The radius of a circle that matches the curve's bend at that specific point; today called the radius of curvature..
If a body were required to move freely along the curve AM with uniform motion, then a normal force would need to be present everywhere pulling the body toward the center MO. This force would have to be in such a proportion to the body’s gravity as the height due to the body's velocity is to half the radius of osculation MO, as is clear from the demonstrations in the preceding Book original: "ex demonstratis Libri praeced. apparet". For unless such a force were present, the body would proceed in a straight line. In this case, however, the channel Euler uses the term "canalis" (channel or tube) to represent any constraint that keeps the body on the path. AM in which the body is conceived to be enclosed prevents the body from proceeding in a straight line. For this reason, the body will press against the channel normally with such a force, in the direction Mn. For if such a normal force were provided by an external source, the body would move freely within the channel AM and would not press against it; but since this external force is absent, as we assume here, it is necessary that the body itself press against the channel with that same amount of force. Which was to be demonstrated.
21. If, therefore, the height due to the body's velocity is set as v and the radius of osculation MO = r, and the gravity of the body is set to $1$ (the weight it would have if placed on the surface of the earth), then the force with which...