Andy: If the answer is somewhere out on the lake and if it is over 12 miles out, it will be invisible, then the height in the image should be 0 inches tall - you can't see it. If, on the other hand, it is closer than the magical 12 miles ( adjustments will have to be made for the curvature effect by cutting off the bottom of the towers appropriately ) then the calculation will still stand. Take the picture, mark the left and right edges of the frame and measure that. Follow the procedure for scaling as if the towers were to be on the shore line. Then, measure the distance from the camera to the shore and use that as a second scaling factor for moving the towers off-shore. Assuming a 1/4 mile width and a 1/4 mile distance from camera to shore, The tower images should be reduced in proportion to the number of 1/4 mile increments off-shore - if the towers are to be 2 mile off shore, the reduction will be 1/8th of the shore height so the towers in my example will be about 3/8ths inches high in a 8 x 12 inch picture, and so on. Without doing the curvature calculations, one can fudge the amount needed to be cut off from the bottom of the towers as a ratio between the 12 mile invisible and actual distance off shore. In the 2 miles off shore example, cut 2/12 or 1/6th off the bottom of the tower and place them exactly on the horizon for an approximate view. ( If any of you out there have the actual formula for calculating the curvature drop-off, please feel free to chip in but I think for this exercise, the linear model will be just fine. ) Cheers, James -----Original Message----- From: owner-photoforum@xxxxxxxxxxxxxxxxxx [mailto:owner-photoforum@xxxxxxxxxxxxxxxxxx] On Behalf Of ADavidhazy Sent: Sunday, March 21, 2010 11:04 AM To: List for Photo/Imaging Educators - Professionals - Students Subject: Re: How tall should the turbines be? I think this is a good solution although in this case there would be no landmarks since the towers are planned for somewhere in a big lake (50 miles wide) - but I think this could be worked out on a map knowing the angle of view of the camera and its location on shore. I was also thinking that since one knows turbine height and distance and the lens focal length one can determine the size of the image of the turbine on the sensor. This size then would bear a relationship to the height of the sensor and this then could be maintained in any final "seascape" illustration. But if the distance is over 3 miles or so then the curvature of the Earth may play a part as well as to how much of the towers reaches above the horizon. If the towers are more than about 12 miles out they would be invisible from eye level on the shoreline. and Emily: >Are these anti-turbine people? If so, are you sure you want to help >them? I'd consider the ethics of this before going any further. I don't know - the person who asked is a reporter for a local newspaper. Andy James Schenken wrote: > The straightforward approach is the take the image of the site and carefully > measure the actual width on the ground from edge of the frame to the edge of > the frame. > For example, suppose the site being pictured is 1/4 mile wide ( 1760 feet ). > A wind turbine sited there will be about 1/4 (400/1760) of the width of the > scene. If the print is 8" by 12" ( assuming a 35mm frame proportions in the > original image ) then the turbine images will be the same ratio in height, > or just under 3 inches high when placed in the image. > Then you can scale the image to whatever final size is needed. > Cheers, > James
