View Full Version : 1 580 = 2 430's
Matt30D
15th of January 2009 (Thu), 21:51
I have a 580 as my master and I want to use my two 430's as slaves. I want to use them for some dance floor shots at the reception....is there a special way to position them ( I will have them on stands). Anyone every try this?
red_click
15th of January 2009 (Thu), 23:13
they'll need to be at least above head height to have even a fighting chance of the E-TTL working as wireless.
Unknowns like disco-lighting etc... would have an impact I guess.
Cant imagine it being reliable :(
Get some radio triggers :)
Wilt
17th of January 2009 (Sat), 11:14
The 580 is brighter than the 430, and you do not normally want your Main light at camera position! So you need to ratio the lights so that the 580 at the camera puts out LESS light than the 430, first of all. If you put both 430's at the same position, and set the 580 to output -1EV compared to the slave, you would have the right balance.
Tim S
17th of January 2009 (Sat), 11:47
The 580 is brighter than the 430, and you do not normally want your Main light at camera position! So you need to ratio the lights so that the 580 at the camera puts out LESS light than the 430, first of all. If you put both 430's at the same position, and set the 580 to output -1EV compared to the slave, you would have the right balance.
Using ratio on the 580? How would you set it (ratio setting) 1:2 or 2:1? Or do you mean Manual using flash meter?
Wilt
17th of January 2009 (Sat), 11:59
Using ratio on the 580? How would you set it (ratio setting) 1:2 or 2:1? Or do you mean Manual using flash meter?
I am not a Canon wireless flash user, so I cannot tell you what actual setting to use...the Canon ratio controls are not exactly stated the same as conventional studio light ratio expression, unfortunately. Let's analyze, and you can then turn this into the control expression on the Master...
The 580EX has output of 1.0 unit of light, while each 430EX outputs 0.5 units of light. So 580EX outputss 1 unit of light at full power, while two 430EX units (together at the same position) output 1 total unit of light. So what you want to do is to tell the 580EX to ouput half as much power as your slave (the pair), or -1EV in light output.
Tim S
17th of January 2009 (Sat), 12:03
I am not a Canon wireless flash user, so I cannot tell you what actual setting to use...the Canon ratio controls are not exactly stated the same as conventional studio light ratio expression, unfortunately.
I thought I had read this somewhere, that's why I was/am confused. Experimentation time I guess.
Thanks!
Wilt
17th of January 2009 (Sat), 12:05
Using ratio on the 580? How would you set it (ratio setting) 1:2 or 2:1? Or do you mean Manual using flash meter?
you need to read the Canon instructions, as I am not a Canon owner.
Titus213
17th of January 2009 (Sat), 12:16
The on-board master is A, the slaves B, the ratios are expressed A:B. So you want 1:2, 1:3, etc. The difference I believe is in the values they express. But that should get you the ratios.
Wilt
17th of January 2009 (Sat), 12:24
I just checked the 580EX diocumentation. They express powers not as f/stop differences but as POWER differences... 1/2 power, 1/4 power, 1/8 power. So in the ratio expression used by Canon, -1EV = 1/2 power = 1:2 on the Master control setting
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