Chapter LXXXI: Section 404: 11th Step

(See the Practice in the 1st Example, Sect. 376.)

_Air_-Thermom. +above+ was 56°.
_Air_-Thermom. +below+ was 63.9
—————
Whole Heat 119.9(0 adding a Cypher)
Half Heat 59.95
Standard-Heat 31.24
which deduct; and there ——————
remains each Moiety, 28.71
above the Standard-Heat.

[Sidenote: _12th Step._]

405. 12th Step. (See the Practice in the first Example, Section 377.)

By the fourth Table, find the Expansion of Air, _with_ 28.71, (more than the Standard-Temperature) _on_ Feet 3935, .1 Tenth, gradually, thus:

406.

_First_ _with_ 28° _on_ Feet 3000 = 204.1[131]
900 as 9000 = 612.3
30 3000 = 204.1
5 5000 = 340.1
.1 1000 = 68.0

Note: 1st. The decimal Point in the Answer corresponding to the Place of _Thousands_, in the Question, is to remain, as taken from the Table calculated for thousand Feet, thus: 204.1.

2d. For _Hundreds_ in the Question, remove the decimal Point _one Place_ in the Answer, thus: 612.3 becomes 61.23:

3d. For _Tens_, _two_ Places, thus: 204.1 becomes 2.041:

4th. For _Units_, _three_ Places, thus: 340.1 becomes .3401:

5th. And for each _Decimal_, a Place more, by adding Cyphers to the left, if wanted, thus: 68.0 becomes .00680.

407. Place the plain and decimated Answers, in one View, and add the latter together, thus:

204.1 = the same 204.1
612.3 = becomes 61.23
204.1 = 2.041
340.1 = .3401
68.0 = .00680
—————————
viz. Expansion of Air _with_} 267.7|179
28° _on_ 3935.1 }

408. _Second_, _with_ .71° _on_ Feet 3000 = 517.5
900 as 9000 = 1552.7
30 3000 = 517.5
5 5000 = 862.6
.1 1000 = 172.5

In order to decimate these Answers, it must be observed that the
Expansion was not _with_ 71 Degrees, but with .71 _Tenths_ of a Degree
of Heat; therefore the decimal Point corresponding to 3000 Feet in the
Question, must in the Answer be removed _two_ Places to the left, thus:
517.5 becomes 5.175: for the 100, three Places: for
1.5527 the 10, _four_ Places: and so
.05175 on.
.008626
.0001725
——————————
6.7|882485

The Expansion with .71 being found, viz.
Feet 6.7 Tenths; add it to the Expansion on
28 Feet already found, viz.
267.7
—————
274.4 Answer.

Which _Height_ in Feet and Tenths, corresponding to the _Expansion_ of Air with 28°.71 Tenths of a Degree of Heat more than the Standard 31°.24, being added to the _Height_ in Feet and Tenths, corresponding to the _Expansion on Inches_ of the Quicksilver in the _upper_ Barometer, with the Standard-Heat, already found, viz. 3935.1 gives the _real Height_ of the _Mountain_, 274.4 —————— or _upper Station_, sought. 4209.5

END OF THE THIRD STAGE.

* * * * *

_The second Example_ briefly _stated: referring to the Sections._

[Sidenote: Section, 391.]

409. Below: Barometer 28.1318.

Attached Thermometer 61°.8; Air ditto 63.9.

Above: Barom. 24.178.

Attached Thermometer 57°.2; Air ditto 56°. Degrees of Heat, viz. 4°.6
to be added to the _colder_ Barometer at Inches 24.178 Tenths, by the
first Table, viz. .0112
Parts of an Inch of the Quicksilver
in the Barometer,
raised by 4°.6 of Heat. ———————
The Sum 24.1892

is the +point+, in Inches and Tenths of an Inch, at which the upper Barometer _now_ rests, being of _equal_ Heat with the lower.

_End of the first Stage._

[Sidenote: Section, 399.]

By the 2d. Table, find the _Height_, in Feet and Tenths, corresponding to the _said_ +point+ when at the Standard-Heat; gradually, thus: the _Height_ corresponding to Feet 24.1 is 7388.0: then with the Difference 107.9, (rejecting the .9).

[Sidenote: Section, 400.]

Find the Height by the 3d. Table corresponding to
.08 86.0 }
.009 9.7 } = Feet 95.9 Tenths.
.0002 .2 }

Which Height subtract from 7388.0
95.9
——————
And there remains, in Feet, 7292.1

The Height corresponding to Inches 24.1892 Tenths of the _upper_ Barometer, with the Standard Temperature of 31.24; for which sole Purpose the 2d. Table is calculated.

Repeat the last Process with the _lower_ Barometer, resting at 28.1318, gradually, thus:

[Sidenote: Section, 401.]

By the 2d. Table, find the _Height_ corresponding to 28.1, which is 3386.61; then with the Difference 92.6 (rejecting the .6) find the corresponding _Height_, by the 3d. Table for the remaining Tenths or Decimals of an Inch, above 28.1, viz. .03 28.0 } .001 .9 } = Feet 29.6 Tenths. .0008 .7 }

[Sidenote: Section 402.]

Which _Height_ subtract from 3386.6
29.6
——————
And there remains, 3357.0 viz. the _Height_ in Feet
corresponding to Inches 28.1318 Tenths of the lower Barometer, with the
Standard Temperature of 31.24, for which sole Purpose the 2d. Table is
calculated.

[Sidenote: Section 403.]

Subtract the _Height_ in Feet, corresponding to Inches of Quicksilver
in the upper Barometer,
viz. 7292.1 from ditto in _lower_ Barometer,
viz. 3357.0 and there remains the _Height_ in Feet
—————— of the upper Barometer at the Standard-Temperature
viz. 3935.1 of 31.24.

_End of the second Stage._

[Sidenote: Section, 404.]

On which Number of Feet, viz. 3935.1, by the 4th Table, find the
_Height_, with 28°.71 of Heat:
_With_ 28°. _on_ Feet 3935.1 = 267.7 and
_With_ .71 _on_ the same = 6.7
—————
Sum 274.4: which
Height, more than the Standard-Heat,
being _added_ to 3935.1
the Height, with the Standard, ——————
gives the true Height, viz. 4209.5.

_End of the third Stage._