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| 0088 |
Explorations in Turkestan : Expedition of 1904 : vol.2 |
Captions
| [Figure] 470 |
Diagram showing Relation between Erosion and Burial of Abandoned Kurgans. Cross-hatching represents wasted top of Kurgan. |
Citation Information
OCR Text
(1) h = l G − [E (t − l) + At] when t < l (G + E)/(A + E)
After that burial takes place, and the depth to which the top is buried at any time
will be:
(2) d = A[t − l (G + E)/(A + E)] when t > l (G + E)/(A + E)
or the rate of aggradation multiplied by the time since foundation minus the time
that elapsed between then and the beginning of burial.
Changing the equation of obliteration somewhat in form, we get our third
and most important equation.
(3) l = t (A + E)/(G + E) when h = o
which means that on aggrading areas any town, not occupied more than the ratio
(A + E)/(G + E) times the number of centuries since it was founded, has vanished from sight
beneath the aggrading plain. The depth to which the eroded top of its accumula-
tion has been buried can be found from equation (2).
Assuming Professor Pumpelly's values obtained at Anau, we have G = 2 and
A = 0.8, and since it is from erosion the growth of plains is supplied and since the
areas of erosion and aggradation seem to correspond in a general way and our cul-
ture mounds probably erode as fast as anything, we may for experiment assume
E = A or E = 0.8. Then (A + E)/(G + E) = 1.6/2.8 = 0.57 as a conservative ratio of much less
error than equation E = A, because E partly compensates itself by division.
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