Salusbury, Thomas
,
Mathematical collections and translations (Tome I)
,
1667
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1 - 10
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31 - 40
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101 - 110
111 - 120
121 - 130
131 - 140
141 - 150
151 - 160
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171 - 180
181 - 190
191 - 200
201 - 210
211 - 220
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251 - 260
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<
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040/01/290.jpg
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becauſe the angle B A D, conteined between the vertical lines, is
<
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equal to the difference of the Polar altitudes, it ſhall be 4
<
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gr. </
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<
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>40m.
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which I note here apart; and I finde the chord of it by the Table
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of Arches and Chords, and ſet it down neer unto it, which is 8142
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parts, of which the ſemidiameter A B is 100000. Next, I finde
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the angle B D C with eaſe, for the half of the angle B A D, which
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is 2
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gr. </
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<
s
>20 m.
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emph.end
type
="
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added to a right angle, giveth the angle B D F 92
<
emph
type
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gr.
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<
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20
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m.
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to which adding the angle C D F, which is the diſtance from
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the vertical point of the greateſt altitude of the Star, which here is
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62
<
emph
type
="
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gr. </
s
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<
s
>15 m.
<
emph.end
type
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it giveth us the quantity of the angle B D C,
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154
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grad. </
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>
<
s
>45 min.
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the which I ſet down together with its Sine,
<
lb
/>
taken out of the Table, which is 42657, and under this I note
<
lb
/>
the angle of the Parallax B C D 0
<
emph
type
="
italics
"/>
gr. </
s
>
<
s
>2 m.
<
emph.end
type
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with its Sine 58.
<
lb
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And becauſe in the Triangle B C D, the ſide D B is to the ſide
<
lb
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B C; as the ſine of the oppoſite angle B C D, to the ſine of the
<
lb
/>
oppoſite angle B D C: therefore, if the line B D were 58. B C
<
lb
/>
would be 42657. And becauſe the Chord D B is 8142. of thoſe
<
lb
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parts whereof the ſemidiameter B A is 100000. and we ſeek to
<
lb
/>
know how many of thoſe parts is B C; therefore we will ſay, by
<
lb
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the Golden Rule, if when B D is 58. B G is 42657. in caſe the
<
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ſaid D B were 8142. how much would B C be? </
s
>
<
s
>I multiply the
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lb
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ſecond term by the third, and the product is 347313294. which
<
lb
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ought to be divided by the firſt, namely, by 58. and the quotient
<
lb
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ſhall be the number of the parts of the line B C, whereof the
<
lb
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midiameter A B is 100000. And to know how many
<
lb
/>
ters B A, the ſaid line B C doth contein, it will be neceſſary anew
<
lb
/>
to divide the ſaid quotient ſo found by 100000. and we ſhall have
<
lb
/>
the number oſ ſemidiameters conteined in B G. </
s
>
<
s
>Now the
<
lb
/>
ber 347313294. divided by 58. giveth 5988160 1/4. as here you
<
lb
/>
may
<
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/>
<
arrow.to.target
n
="
table13
"/>
</
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>
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>
<
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>
<
table.target
id
="
table10
"/>
<
row
>
<
cell
/>
<
cell
>
<
emph
type
="
italics
"/>
gr.
<
emph.end
type
="
italics
"/>
</
cell
>
<
cell
>
<
emph
type
="
italics
"/>
m.
<
emph.end
type
="
italics
"/>
</
cell
>
<
cell
>Its chord 8142 of thoſe</
cell
>
</
row
>
<
row
>
<
cell
>Ang. B A D</
cell
>
<
cell
>4</
cell
>
<
cell
>40</
cell
>
<
cell
>parts, whereof the ſemid.</
cell
>
</
row
>
<
row
>
<
cell
>B D F</
cell
>
<
cell
>92</
cell
>
<
cell
>20</
cell
>
<
cell
>A B is an 100000.</
cell
>
</
row
>
</
table
>
<
table
>
<
table.target
id
="
table11
"/>
<
row
>
<
cell
>B D C</
cell
>
<
cell
>154</
cell
>
<
cell
>45</
cell
>
<
cell
>Sines</
cell
>
<
cell
>42657</
cell
>
</
row
>
<
row
>
<
cell
>B C D</
cell
>
<
cell
>0</
cell
>
<
cell
>2</
cell
>
<
cell
/>
<
cell
>58</
cell
>
</
row
>
</
table
>
<
table
>
<
table.target
id
="
table12
"/>
<
row
>
<
cell
>58</
cell
>
<
cell
>42657</
cell
>
<
cell
>8142</
cell
>
</
row
>
<
row
>
<
cell
/>
<
cell
>8142</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>85314</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>170628</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>42657</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>341256</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>59</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
>58</
cell
>
<
cell
>3473</
cell
>
<
cell
>13294</
cell
>
</
row
>
<
row
>
<
cell
/>
<
cell
>571</
cell
>
<
cell
/>
</
row
>
<
row
>
<
cell
/>
<
cell
>5</
cell
>
<
cell
/>
</
row
>
</
table
>
<
table
>
<
table.target
id
="
table13
"/>
<
row
>
<
cell
/>
<
cell
>5988160 1/4</
cell
>
</
row
>
<
row
>
<
cell
>58</
cell
>
<
cell
>347313294</
cell
>
</
row
>
<
row
>
<
cell
/>
<
cell
>5717941</
cell
>
</
row
>
<
row
>
<
cell
/>
<
cell
>543</
cell
>
</
row
>
</
table
>
<
p
type
="
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<
s
>And this divided by 100000. the product is 59
<
lb
/>
<
arrow.to.target
n
="
table14
"/>
</
s
>
</
p
>
<
table
>
<
table.target
id
="
table14
"/>
<
row
>
<
cell
>1 |00000</
cell
>
<
cell
>| 59 |</
cell
>
<
cell
>88160.</
cell
>
</
row
>
</
table
>
<
p
type
="
main
">
<
s
>But we may much abbreviate the operation, dividing the firſt
<
lb
/>
quotient found, that is, 347313294. by the product of the
<
lb
/>
plication of the two numbers 58. and 100000. that </
s
>
</
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>
</
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