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0054 Scientific Results of a Journey in Central Asia, 1899-1902 : vol.1
Scientific Results of a Journey in Central Asia, 1899-1902 : vol.1 / Page 54 (Grayscale High Resolution Image)

Captions

[Figure] Fig. 37 10 April 1900. Tusun-tschapghan.
[Figure] Fig. 38 10 April 1900. Kum-tschapghan.
[Figure] Fig. 39 22 Sept. 1899. Schakal-otak.

New!Citation Information

doi: 10.20676/00000216
Citation Format: Chicago | APA | Harvard | IEEE

OCR Text

 

32

THE TARIM RIVER.

We get by numerical calculation

according to method I according to method II

A   dm   Va

m3

   2 6 0'24 -   7 8.15 m2   2.6o5 m   0.3332

   sec.   sec.

3

   2 6 24 m   7 8.03 m2   2.601 m   0.3363 m

   sec.   sec.

Fig. 37. Left. r o April 1900. b = 22.o metres. n = 11 Tusun-tschapghan. Right.

p . . 1.06 2.87 4.67 4.8o 4.36 3.75 2.41 I.30 0.6z 0.16 metres

v/k . . 35.o 52.o 45.25 41.o 40.o 37.2 30.3 34.5 18.5 28   k second
The results obtained are

according to method I    

according to method II   

Q

m3

19'51 sec.

m3

19.39 sec.

A   dm   Vm

5 2.6o m2   2.391 m   0.3709 m

sec.

5 2.00 m2   • 2.364 m   0.3728 m

sec.

Fig. 38. Left. ro April 190o. b = 22.6 metres. n = 1 r Kum-tschapghan. Right.

   p . . 0.46 0.88 I.16 I.88 I.48 0.86 1.63 I.99   1.96 I.37 metres

v/k .   2.00   8.o 28.7 26.3 19.5   I.   2.   metre

3•   9.5 4 7 47•0 5 3 45.7 second

Here the results are

Q   A   dm   Vm

m3

according to method I   8 52 -   2 7.86 m2   1.233 m0.3058 m

   sec.   sec.

according to method II   8 58 m3   28.o9 m2   1.243 m   0.3055 m

   sec.   sec.

Fig. 39. Left. 22 Sept. 1899. h = 35.3 metres. n = 6 Schakal-otak. Right.

ft    3.72   4.98 4.77   3.09   1.23 metres

v/k . . .     83.o 78.5 71.o 53.5   4.5 1 metre

k second

And the results work out thus

Q   A   dm

according to method I   68 30 m3   109.03 m2   3.089 m

sec.

according to method II   66 16 m3sec.   104.66 m2   2.965 m

0.6264 m

sec.

0.6321 -

sec.