 BetterPhoto Member |
First Wedding Shoot Well, I have finally/possibly landed a wedding, and of course I have absolutley no idea how much to charge. So Help. They need me for a couple hours just doing the portraits and thats it.
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Jerry Frazier |
$6,000
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W. |
"They need me for a couple hours just doing the portraits and thats it." What? No ceremony? No reception? No party?
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Mark Feldstein |
Well, first Andi, it depends on what pixels are selling for these days. Hey Jer, how much is a pound of pixels lately? WS may know too. Or just tell them you'll bill for parts, no labor after you see if and how they turn out. I mean, you DO have some experience photographing for money, don't you? Along with all the usual annoying things photographers have to have like,ohhhhhhhhhhhhhhhh saaaayyyyy liability insurance, equipment depreciation schedules, quarterly IRS returns, state tax, a lab fee list of costs and examples, album costs, pixels, spare cards, flash units, assistant if any, transportation costs are figured in at ___ per mile, pre and post shoot time, and all that stuff. You've got that all figured out, right Andi? Yes??? Right? I think 6K is a tad low, but ok if you're only giving them 3x5 post card prints stamped SAMPLE.
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Pete H |
The following is my exhaustive list of what you need to do Andi; based only on your question.
Pete
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Pete H |
Oh sorry, you wanted an answer; right? Ok..This is the basic formula most of us who know what we are doing use. First, figure the std deviation of cost VS material; but only under the bell curve.. Go to -1 on both sides, disregarding any fliers...use the french curve rather than integrating. If you are using a digital camera, you will have to sample the square root rather than average. .707 should allow you to filter to a RMS value. Usually a rate of (100Hz/shutter) speed is fine. Ok; next, summate all primariy factors; unless it's Thursday, in which case you need to find the tangent of 45 degrees..usually one; unless you are using a high mega pixel camera. Finally sum the angles of all your lenses, (IS) notwithstanding. Any values greater than 1 need to be discarded or squared under the rule of nines; unless you are using a fast lens like f/1.8..In that case you will have to consult a book on modern cosmology. Don't worry, I've never had to do that yet, but like you I am willing to learn. Make sure the (-sin) of all your angles are inclusive before presenting numbers to your client. Use (cosin) if shooting film. I know as seasoned veterans, Jerry, W.S and Mark will agree whole heartedly, this is the way to price your time as a photographer. all the best, Pete
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Todd Bennett |
Dang, I thought you could charge $75.00 per hour and be cool!
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Mark Feldstein |
Well Todd, I think Pete kind of hit it with RMS value and .707 filter factor although I might add subtracting that from the hyperfocal distance of either 4x5" converted to centimeters with a dimpling pin hypertheoretical replacement coefficient of 6,02x10 to the 23rd power. Just an alternate way of calculating that part of the formula. BUT if it's on a Saturday, there's no parking of course on Broadway between Union Square and you have to factor in Bart Fares. Just thought I'd mention that. But yes, great minds think alike and I concur in this diagnostic extravaganza and misadventure.
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