Calculation of the average angle of the KCN and KCS shafts by the known positions of the inlet (inside the chamber) and outlet for each shaft.




Northern shaft.

Input data:
1) FH = 79.24m [3119.8 in.; Petrie, Pl.XI] or 80.63m [154c of 0.5236m/c; Gantenbrink] - elevation of the KCN shaft outlet (to the restored casing);
2) EH = 43.91m – elevation of the KCN shaft inlet (elevation of the KC floor + elevation of the shaft inlets above the KC floor = 42.99m [1692,5 in. (avg.); Petrie, sect.55] + 0.92m [Maragioglio, Rinaldi, Tav.7 Fig.1]);
3) KL = 230.35m [9069 in. (avg.); Petrie, sect.21] – casing side length;
4) DM = 5.77m – distance from north end of KCN horizontal section to centre (distance from KC north wall to centre - length of KCN horizontal section = 8.40m [330.6 in. (avg.); Petrie, sect.55] - 2.63m [Gantenbrink, [www.cheops.org]]);
5) <FLH = 51° 52’ (Petrie, pl.XI).

<FDE = atan(FE/DE);
FE = FH - EH;
DE = ME + DM = NL - HL + DM;
NL = KL/2;
HL = FH/tan(<FLH);
<FDE = atan((FH - EH)/(KL/2 - FH/tan(<FLH) + DM));

Petrie: <FDE = atan((79.24 - 43.91)/(230.35/2 - 79.24/tan(51° 52’) + 5.77)) = 31.026° or 31° 01’ 34’’ (or ~11 palms 3 digits = 30° 47’ 03’’; the difference is 14’ 31’’).

Gantenbrink: <FDE = atan((80.63 - 43.91)/(230.35/2 - 80.63/tan(51° 52’) + 5.77)) = 32.496° or 32° 29’ 47’’ (or 11 palms = 32° 28’ 16’’; the difference is 01’ 31’’).


Southern shaft.

Input data:
1) AG = 79.88m [3144.9 in. (calculated; see notes); Petrie, pl.XI] or 80.63m [154c of 0.5236m/c; Gantenbrink] - elevation of the KCS shaft outlet (to the restored casing);
2) BG = 43.91m – elevation of the KCS shaft inlet (elevation of the KC floor + elevation of the shaft inlets above the KC floor = 42.99m [1692,5 in. (avg.); Petrie, sect.55] + 0.92m [Maragioglio, Rinaldi, Tav.7 Fig.1]);
3) KL = 230.35m [9069 in. (avg.); Petrie, sect.21] – casing side length;
4) CM = 15.36m – distance from south end of KCS horizontal section to centre (distance from KC south wall to centre + length of KCS horizontal section = 13.64m [537 in. (avg.); Petrie, sect.55] + 1.72m [Gantenbrink, [www.cheops.org]]);
5) <AKG = 51° 52’ (Petrie, pl.XI).

<ACB = atan(AB/BC);
AB = AG - BG;
BC = KN - KG - CM;
KN = KL/2;
KG = AG/tan(<AKG);
<ACB = atan((AG - BG)/(KL/2 - AG/tan(<AKG) - CM));

Petrie: <ACB = atan((79.88 - 43.91)/(230.35/2 - 79.88/tan(51° 52’) - 15.36)) = 44,109° or 44° 06’ 34’’ (or ~7 palms 1 digit = 43° 59’ 42’’; the difference is 06’ 52’’).

Gantenbrink: <ACB = atan((80.63 - 43.91)/(230.35/2 - 80.63/tan(51° 52’) - 15.36)) = 45.159° or 45° 09’ 32’’ (or ~7 palms = 45° 00’ 00’’; the difference is 09’ 32’’).


Notes:
1) All values and calculations are for the bottom (lower plane) of shafts.
2) The value of 3144.9 in. is calculated according to Petrie's drawing for the KCS shaft outlet to the restored casing plane, pl.XI.


Sources:
1) Petrie, W.M.F. (1883), The Pyramids and Temples of Gizeh, (Field & Tuer, London).
2) Gantenbrink, R. ([www.cheops.org]).
3) Maragioglio V., Rinaldi C. (1965), L'Architettura delle Piramidi Menfite 4. La piramide di Cheope. Tavole, (Rapallo).


Preliminary conclusions:
1) Calculation shows that the average angles of both KC shafts according to Gantenbrink were not actually measured by him, but were calculated and rounded to the nearest whole palm seked to correspond to the “theoretical” elevation of the outlets of 154c. (he used the 154c level as input data and got average angles as output data as we did)
2) Since we know that the both outlets would not actually reach the 154c level, but should exit slightly lower and at different elevations (Petrie is correct in the data on the heights of the exits, see above in the thread), then the actual average angles of both shafts are slightly less than those calculated by Gantenbrink (according to our calculation, the average angle of KCN is less by 1.5°, and the average angle of KCS is less by 1° than Gantenbrink's values).
3) Petrie (sect.56) writes about the average slope of the shafts: "It is striking that the slope of both passages continuously increases up to the outside (except just at the mouth of the S. channel); hence these quantities, which only extend over a part of either passage, cannot give the true mean slope; probably on the whole length the means would not be greater angles than 31º and 44½º respectively." His expectations for KCN exactly correspond to our calculation, and for KCS differ by only 23'.

I would be grateful if someone would check the correctness of this simple calculation so that final conclusions can be drawn.

Alex.



Edited 1 time(s). Last edit at 11/16/2020 06:57AM by keeperzz.