 | Jeremiah Day - Measurement - 1815 - 388 pages
...equation above will then become Dxh=d3. Therefore h- — That is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex. 1. What is the difference between the true and the apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter... | |
 | Jeremiah Day - Geometry - 1824 - 440 pages
...equation above will then become Dxh=d3. Therefore A=^ That is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex. 1- What is the difference between the true and the apparent level, fora distance of one English mile, supposing the earth to be 7940 miles in diameter?... | |
 | Jeremiah Day - Measurement - 1831 - 394 pages
...above will then become > d» Dx.h=d*. Therefore A=-p. That is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex. 1. What is the difference between the true and the apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter... | |
 | Jeremiah Day - Logarithms - 1831 - 418 pages
...equation above will then become d3 Z>. Therefore h=-. That is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex. 1. What is the difference between the true and the apparent level, for a distance of one English rnile, supposing the earth lo be 7940 miles in diameter?... | |
 | Jeremiah Day - Geometry - 1838 - 416 pages
...equation above will then become d2 DxA=d2. Therefore A=gThat is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex. 1. What is the difference between the true and the apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter?... | |
 | Jeremiah Day - Geometry - 1839 - 434 pages
...equation above will then become Dxh=d2. Therefore A ==rThat is, the difference beticeen the trye and the apparent level, is nearly equal to the square of the...Ex. 1. What is the difference between the true and the apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter... | |
 | Charles Guilford Burnham - Arithmetic - 1841 - 324 pages
...true level, at the point where the observation is made. 3. The difference between the true and the apparent level is nearly equal to the square of the distance, divided by the diameter of the earth. 1. What is the difference between the true and apparent level, for a distance of one English mile,... | |
 | Charles Guilford Burnham - 1850 - 350 pages
...true level, at the point where the observation is made. 3. The dilference between the true and the apparent level is nearly equal to the square of the distance, divided by the diameter of the earth. 1. What is the difference between the true and apparent level, for a distance of one English mile,... | |
 | Jeremiah Day - Geometry - 1851 - 418 pages
...equation above will then become d2 DxA=d2. Therefore A=gThat is, the difference between the true and the apparent level, is nearly equal to the square of the...Ex: 1. What is the difference between the true and the apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter... | |
 | William Smyth - Navigation - 1855 - 234 pages
...whence ™ = £^. That is, the difference between the apparent and true level for any distance, is equal to the square of the distance divided by the diameter of the earth. Ex. Let AD = a statute mile, or 5280 feet ; and 2 CD, the diameter of the earth, = 7912 miles ; we have... | |
| |
A=gThat is, the difference between the true and the apparent level, is nearly equal to the square of the distance divided by the diameter of the earth. Ex. 1. What is the difference between the true and the apparent level, for a distance of one English... 
