CHAPTERLXXIX.A CALCULATION TO ASCERTAIN THE HEIGHT OF THE BALLOON ON THE DAY OF ASCENT: ONE BAROMETER AND ONE THERMOMETER ONLY, BEING TAKEN UP INTO THE CAR.Section 423.THE Question is stated from Section 36: and the Mode of Operation taken from theRecapitulationof the second Example, Section 409.Observation before the Ascent:Below: Barometer 29.8; attached Thermometer 0; detached Thermometer 65°.Above: Barometer 231⁄4 = 2325⁄100 or 23.25[135]attached Thermom. 0; detached Thermom. 65°.There being no attached Thermometers; thefirstTable is useless: the Barometer below is therefore supposed to be of the same Temperature as when above; the detached Thermometer remaining at the same Degree, viz. 65°.State the Barometer, thus: whenbelow, at29.8whenabove, at23.25.End of the first Stage.424. Find the Height (at the Standard-Heat) corresponding to the Inches andnearestTenth above and below 23.25: i. e. above 23.2, and below 23.3: by the 2d Table.Now 23.2 corresponds to 8379.7: and the Difference of .1 above, i. e. to 23.3, is in Feet = 112|.1: by the 3d Column of the same Table.With this Difference, consult the 3d Table: i. e. with 112, (omitting the .1 as too minute) on the remaining Decimals above 23.2, viz. on 05, as on 5, or 5⁄10; and the Answer is 56 Feet: which Number being subtracted from 8379.7, the Remainder 8323.7, is the Height in Feet of the Barometer in the Car, at the Standard-Heat.Repeat the last Process for the Barometer on the Ground.Now 29.8, by the 2d Table, corresponds to 1856.0; and there being no Parts or Decimals more minute than a Tenth, viz. .8, there is no Occasion for the 3d Table.Subtract the Barometer in the Car, from the same when on the Ground; and, by the 2d Table,upper Barom.23.25,corresp. to8323.7,and thelower Barom.29.8,to1856.0:theRemainder is the Height in Feet———of theBarometer in the Carviz.6467.7,with the Standard-Heat.End of the second Stage.425. Detached Therm. above, at65°Detached Therm. when below, at65——Whole Heat130Half Heat65.(00 adding Cyphers)Standard-Heat31.24——which deduct, and there remains33.76Degreesmore than the Standard-Heat, for each Barometer.Then for the Expansion of Air, with such Heat more than the Standard, consult the 4thTable: viz.with33°.76onInches 6467.7, the Height of the Barometer in the Car with the Standard-Heat, thus:426.First,with33°,on6467.7on6000 as 6000 = 481.1,decimated481.1400 as 4000 = 320.732.0760 as 6000 = 481.14.8117 as 7000 = 561.3.5613.07 as 7000 = 561.3.05613————Expansion =518.59843427.Second,with.76on6467.7:on, as before,6000 = 1108.decim.11.084000 = 738.7.73876000 = 1108..11087000 = 1292.7.0129277000 = 1292.7.0012927—————Expansion =11.9437197Add the former518.59843—————Total Expansion =530.5|542197viz. Heightby Expansionin Feet,with more than the Standard-Heat,add to Height in Feet atthe Standard-Heat6467.7428. The true Height, in Feet andTenths, of the Barometer in theCar6998.2Feet in a Yard 3)———Yards in a Mile 1760)2332.2Feet.1760(1 Mile.———Yards in a Quarter of a Mile 440)572(1 Qr.440——32Yards.The Height of the Balloon 1 Mile, 1 Quarter, 32 Yards, and 2 Feet.End of the last Stage,and of the Mensuration of Heights.N. B. Athermometricsliding Rule, for the Expansion of Quicksilver, and of Air, may possibly, from the foregoing Tables, be so contrived and adapted to the Barometer, as to tell the Height by Inspection, while in the Car of the Balloon.
CHAPTERLXXIX.
Section 423.THE Question is stated from Section 36: and the Mode of Operation taken from theRecapitulationof the second Example, Section 409.
Observation before the Ascent:
Below: Barometer 29.8; attached Thermometer 0; detached Thermometer 65°.
Above: Barometer 231⁄4 = 2325⁄100 or 23.25[135]attached Thermom. 0; detached Thermom. 65°.
There being no attached Thermometers; thefirstTable is useless: the Barometer below is therefore supposed to be of the same Temperature as when above; the detached Thermometer remaining at the same Degree, viz. 65°.
State the Barometer, thus: whenbelow, at
whenabove, at
End of the first Stage.
424. Find the Height (at the Standard-Heat) corresponding to the Inches andnearestTenth above and below 23.25: i. e. above 23.2, and below 23.3: by the 2d Table.
Now 23.2 corresponds to 8379.7: and the Difference of .1 above, i. e. to 23.3, is in Feet = 112|.1: by the 3d Column of the same Table.
With this Difference, consult the 3d Table: i. e. with 112, (omitting the .1 as too minute) on the remaining Decimals above 23.2, viz. on 05, as on 5, or 5⁄10; and the Answer is 56 Feet: which Number being subtracted from 8379.7, the Remainder 8323.7, is the Height in Feet of the Barometer in the Car, at the Standard-Heat.
Repeat the last Process for the Barometer on the Ground.
Now 29.8, by the 2d Table, corresponds to 1856.0; and there being no Parts or Decimals more minute than a Tenth, viz. .8, there is no Occasion for the 3d Table.
Subtract the Barometer in the Car, from the same when on the Ground; and, by the 2d Table,
corresp. to
to
viz.
End of the second Stage.
65°
65
——
130
65.
31.24
——
33.76
Then for the Expansion of Air, with such Heat more than the Standard, consult the 4thTable: viz.with33°.76onInches 6467.7, the Height of the Barometer in the Car with the Standard-Heat, thus:
426.First,with33°,on6467.7
6000 as 6000 = 481.1,
481.1
400 as 4000 = 320.7
32.07
60 as 6000 = 481.1
4.811
7 as 7000 = 561.3
.5613
.07 as 7000 = 561.3
.05613
————
518.59843
427.Second,with.76on6467.7:
6000 = 1108.
11.08
4000 = 738.7
.7387
6000 = 1108.
.1108
7000 = 1292.7
.012927
7000 = 1292.7
.0012927
—————
Expansion =
11.9437197
Add the former
518.59843
—————
Total Expansion =
530.5|542197
6467.7
6998.2
Feet in a Yard 3)
———
Yards in a Mile 1760)
2332.2
1760
———
Yards in a Quarter of a Mile 440)
572
440
——
32
The Height of the Balloon 1 Mile, 1 Quarter, 32 Yards, and 2 Feet.
End of the last Stage,and of the Mensuration of Heights.
N. B. Athermometricsliding Rule, for the Expansion of Quicksilver, and of Air, may possibly, from the foregoing Tables, be so contrived and adapted to the Barometer, as to tell the Height by Inspection, while in the Car of the Balloon.