[ckfong2002]

先假設兩枝燈同主體一樣距離, 4 米.

如果主燈GN32, 副燈GN24, 一齊閃, GN 點計呢?

定還是只可以用測光錶計番轉頭?

[hktsoi]

GN40

[香港至NET超人]

Let Nth number of flash with GN from N1 to Nn:

The effective GN = (N1^2 + N2^2 + N3^2 + ...... + Nn^2) ^(1/n)

This is just the assumption on all flash pack closely and point to the same target point.

[andrew]

照超人的 formula 應該是 (32^2 + 24^2)^1/2 = 26 GN. 對不對?

[ChaseWind]

(32^2 + 24^2)^1/2 = 40
andrew哥哥計錯數哩^^

[ckfong2002]

唔該晒呀, 你地回文好快. :)

[andrew]

係喎... 按錯 calculator 添... :P

[rencontre]

唔明wor,
' ^ '呢個符號係乜 ??

[Velsper]

^
"次方"
2^3
"2乘以三次方"
= 8

[Velsper]

補充:
2乘以三次方 = 2 x 2 x 2
= 8

[rencontre]

明la,
thanks,

咁如果3支外閃又點計 ??

[rencontre]

係咪:

(n1^2+n2^2+n3^2)^1/3 ???

[rencontre]

仲有, 如果三支燈`距離唔同,
計法又如何??

[hkhei]

>(n1^2+n2^2+n3^2)^1/3 ???
唔係咁計
overall gn=(gn1^2+gn2^2+gn3^2+.....)^1/2

proof:
因為gn 值proportional to effective range r
所以gn=k1*r
而output porportional to gn^2
ie output =k2*gn^2

total output =output1+output2 + output3+....
ie total output = k3*(gn1^2+gn2^2+gn3^2+...)
但係output =k2*gn^2
即係話overall gn^2=k4*(gn1^2+gn2^2+gn3^2+...)
overall gn^2=k5*(gn1^2+gn2^2+gn3^2+...)^1/2
where k1,k2,k3,k4,k5 are constant...
事實上，k5=1
所以overall gn^2=(gn1^2+gn2^2+gn3^2+...)^1/2

[hkhei]

>仲有, 如果三支燈`距離唔同,
呢個要睇三支燈點放法....