stands, is always in the surface of the surrounding water, since the surface vessel does not weigh) be equal in weight to the given body A B, Reason, that two weights that cause a vessel to sink equally deep are also equal in weight, by the 3rd postulate. CONCLUSION. A solid body then lighter than water that lies therein, is equal in weight to the water equal in size to its part that is within the water, which we had to prove.

A geometric diagram shows a rectangular body labeled A, B, and E, which is partially submerged within a larger rectangular frame representing water or a container, labeled C and D.

I DEMAND. VI PROPOSITION.

ONE PART OF THE SOLID BODY OF KNOWN SIZE LYING IN WATER OF KNOWN WEIGHT, AND THE OTHER PART OUTSIDE IT: TO FIND THE WEIGHT OF THE WHOLE BODY.

GIVEN. Let A B C D be a solid body of form the text implies a specific shape, likely a ship and E F a water of which a teerlincksche voet cubic foot weighs 65 lb (so much weighs according to experience a Delft foot of Delft water, and thereon we shall always estimate in the following examples) and the body's part within the water be A C D, whose size be of 10,000 cubic feet. REQUIRED. We must find how heavy the whole body A B C D is, with all that is in and upon it.

WORK. One shall multiply 10,000 by the 65 lb, comes 650,000 lb for the required.

PROOF. The whole body A B C D is equal in weight to the water equal in size to A C D by the 5th proposition, but the water equal in size to A C D weighs 650,000 lb, the whole body then A B C D weighs 650,000 lb, which we had to prove.

An illustration shows a three-masted sailing ship on the water. The ship's hull is labeled A, B, C, and D. The water surface level is indicated by a line labeled E and F. The ship is shown within a rectangular frame that represents the water depth.

CONCLUSION. One part of the solid body of known size lying in water of known weight, and the other part outside it; we have found the weight of the whole body according to the demand.