150 UN NUOVO METODO PER DETERMINARE LA RESISTENZA DELL'ARIA 
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12 I 24 36 
x ettometri | 
n =4,16241 H = 21,306 Nel cgli 2,21788 |3,43576 4,65364 
logn —=0,6193448 logH 1,3285019 ‘121788 (1,21788. | 
tag (1 RE O leale |0,3459382|0,5360228|0,6677928 
login—2)=0,3349380 log (n—2) =0,3349380 —l0gN......-.... ,3459382|0, 8000 
log2 = =03010300 loge(12) =1,0791812  log(logM...... ‘9,539898419,7291832/9,8246417 
log = 9,9660920 log? = 0,0856173. log— ....... 9,9660920/9,9660920|9,9660920 
n_-2 H n DÀ 
x =1,21788 log (—— og N) |[9,5059904/9,6952752/9,7907337 
n= 2 
9 
— log N . |0,3199562/0,4957642|0,6176476 
long Souaai) 7,1873365|7,7873365|7,7873365 
logkN®=3...... 8,1072927 8,2831007|8,4049841 
Gb LI 0,012802 |0,019191 |0,025409 
pi Vi 277 | 226 197 
2 | 
ClogH = 8,6714981 zlogN (0,31995620,4957642|0,6176476 
log 2 = 0,3010300 si 
log (n—1) = 0,5000182 21og N 0,6918764 /1,0720456|1,3355856 
log (r=12) =1,0791812 CORR n 
— n — Ì - 
De = 5 1,0118326|1,5678098|1,953 
loge"= 2 3 = 0,5517275 0 log N “ logN+2logN |1,0 oi 5) 1,9532332 
i — 3,56278 Na 10,27620 |36,96663 189,79109 
log = 7,7873365 PAPA Liu 4,56228 18,12456 11,68684 
2log H = 2,6570038 H 3,56228 |3,56228 
Clogn = 9,3806552 palati Q2n—2 
Clog m—1) =9/4999818 N3->_1- 2 [571492 |18,84207 |78,10425 
kH° auni 2n—2 
log—-— = 9,3249773 log (x mail pil 7) ‘0,7569341 anti 
n(n—1) H I | 
1) | | 
l087T=1) (9,3249773 9,3249773|9,3249773 
Clog x (8,9208188 8,6197888 |8,436975 
s Sace i_QIGA VA | 
logsen?2 p=log ca ( 1 ca 2 x) [9,0027302 9,4047927/9,6613495 
2°,53',3 7°,21', 4 |13°,38',7 
logH = =1,3285019 cir 01% 0,5059163|0,7859049|0,9766166 
Clog V = 9,3979400 (A | 
Clog (ni) .=9,4999818 Nes 320565 /6,08002 [9,4175815 
log EU FE de 0,2264237 log ( Wi-A) 0,3435367/0,7058654|0,9281815 
0,2264237|0,2264237|0,2264237 
0,0005528 0,0035914/0,0124430 
) |0,5705132/0,9358805|1,1670482 
H 
(n 1)V 
3", 7I9| 8”, 627 wr, 691 
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