 | Albert Edward Seaton - Marine engineering - 1886 - 520 pages
...path so found is like an attenuated figure 8 in head-gear, and somewhat more pronounced in sterngear. The arc of a circle of radius equal to the length of the suspension or bridle rods, is then drawn through each of these figures in such a way that there... | |
 | Albert Edward Seaton - Marine engineering - 1886 - 514 pages
...path so found is like an attenuated figure 8 in head-gear, and somewhat more pronounced in sterngear. The arc of a circle of radius equal to the length of the suspension or bridle rods, is then drawn through each of these figures in such a way that there... | |
 | Emory Edwards - Marine engineering - 1891 - 554 pages
...path so found is like an attenuated figure 8 in head-gear, and somewhat more pronounced in stern-gear. The arc of a circle of radius equal to the length of the suspension or bridle rods, is then drawn through each of these figures in such a way that there... | |
 | 1902 - 486 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | American School (Chicago, Ill.) - Engineering - 1903 - 414 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | American School (Lansing, Ill.) - Architectural drawing - 1906 - 426 pages
...at the same distance from the vertfix. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | American School (Chicago, Ill.) - Engineering - 1906 - 580 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | Drawing - 1906 - 424 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | American School (Lansing, Ill.) - Architecture - 1907 - 360 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
 | American School of Correspondence - Civil engineering - 1908 - 474 pages
...at the same distance from the vertex. From this it follows that the development of the base will be the arc of a circle of radius equal to the length of an element. To find the length of this arc which is equal to the distance around the base, divide the plan of the... | |
| |
... element is taken from the elevation A, of either contour element. All of the elements of the cone are of the same length, so when the cone is rolled each point of the base as it touches the plane will be at the same distance from the vertex. From... 
