EX1 sensor dimensions?
 

Has anyone come across information about the actual dimensions (mm) of the EX1's imaging area? I'd like to calculate DOF and hyperfocal distances before the camera gets here. To do that I need to know the physical dimensions of the image at the chips.
Thanks.
Clark

I don't know the sensor size but i think you can work out some things based on the FOV multiplication factor because you know the widest lens and what it's 35mm equivalent is from specs (5.8mm is 31.4mm on 35)

So for example (please shoot me down if im wrong)

the crop factor for the EX is 5.41, the HVX is 7.73 and my EOS 300 is 1.6.

So if the EX was f1.9, that would be f10 on 35mm and f6.4 on the eos and the HVX couldn't open that wide. If the HVX was on f2 that would be around f2.9 on the EX, about a stop which is what others have been saying.

I'm pretty sure aperture is independent to focal length but obviously the magnification of the DOF blur would increase with longer focal lengths (which makes it look like you get more DOF with a long lens)

Can someone back me up here :)

paul

the diagonal size of a CMOS sensor on a 1/2" optical system should be 8mm based upon the formula diag/16=lens format. 8/16=1/2.

that's how i read it. now you have to do the rest of the math to get your HxW.

Vince

I tried your ideas, but...
 

"the diagonal size of a CMOS sensor on a 1/2" optical system should be 8mm based upon the formula diag/16=lens format. 8/16=1/2."

I don't know the source of this formula. I tried it and came up with a sensor size of 7mm wide by 4 mm high. The diagonal of 8mm is woefully short of 1/2". I don't think this works.

I tried the crop factor method suggested by Paul. This is a little iffey because I don't know how Sony compared the EX1 to 35mm. The proportions don't match. I took a try comparing the diagonal measures of the two formats. I ended up with a sensor size of 9.1mm by 5.1mm and a diagonal of 10.4mm. This is closer to 1/2", but it's still just a shot in the dark. Lacking confirmed actual measurements I guess I could do my DOF and hyperfocal calculations based on this sensor size. It's less precise than I'd like to be, but nothing about DOF is precise.

Thanks.
Clark

Film apertures
 

A chance to dust off my American Cinematographer Manual that I never found a use for in school!
According to the manual, a 35mm Academy aperture is 0.868 inches by 0.631 inches. Of course, that's for a 4:3 aspect ratio.
(I knew there was a reason why I was keeping that old book around!)

But these 35mm equivalent focal lengths might be for full-frame still cameras, in which case I measure the aperture as 36mm by 24mm, or about an inch and a half by an inch.

In any case, I'll leave it to someone else to calculate the DOF tables. :)
Ken Hull

i believe that 1/2" refers to the lens not the imager. they call it a 1/2 type exmor cmos. type referring to the optical format which is based on the lens family.

also, if you take the 5.41 crop factor from paul and divide that into 43.2 which is the diagonal measurement of a 35mm still frame it comes out to just about 8.

hey, it seems really small to me too.

vince

here it is:

The actual image size (active area) = 6.4mm x 4.8mm with the image diagonal = 8mm. The 1/2 inch (1/2” = 12.70mm) value that is used to describe the common format name stems from the days of Videcon Tubes and was used to describe the tube diameter and not the actual image size.

From what I've read, a 1/2" sensor has essentially the same DOF characteristics of 8mm film. So the actual dimensions of an 8mm negative should be pretty darn close to the EX-1's sensor size.

Assuming that the sensor diagonal is 1/2 inch (12.7 mm) and the aspect ratio is 16:9, Pythagoras gives the dimensions as 6.23mm x 11.07 mm.

Nick

And that is much closer to Super 16 mm film (7.4 x 12.4 mm) than to Super 8 (4.2 x 5.7 mm).

I don't think it is as simple as Pythagorus could make this, although I wish that it was. I read somewhere on these boards(I think it was Barry Green), that the sensor size is actually quite a bit smaller than the actual chip. They were discussing how the dimension is actually quite a bit decieving to say 1/2" or 1/3" because the original dimension had something to do with a tube size in old cameras. I can't remember the details other than the summation that the chip size may be 1/2 inch, but the actual sensor on the chip is much smaller than that. I definitely could be wrong.

http://de.wikipedia.org/wiki/16_mm

Der 16 mm-Film verfügt über eine Bildfeldgröße von 10,3 x 7,5 mm (Breite x Höhe) und kann sowohl einseitig als auch zweiseitig perforiert genutzt werden, die Bilddiagonale beträgt also 1/2 inch (Zoll) entsprechend 12,7 mm.

but this is seen as letterbox. Because of 16:9 the 4:3 equivelent would be a bit bigger. On the DOF thread somebody has posted a nice image with relatively shallow DOF on a normal focal length (in-room), so the EX-1 should do well for storytelling.

Ah good lord you guys are making my head hurt. Just put the damn thing in auto focus, or just eyeball it. That should be good enough.
Sarcasm aside, reading these message boards constantly keeps me amazed at how much I don't know about my craft... and never knew that I didn't know.

Hey, we got snowed in today. I just googled till I found the size. And realized when I did find it, I don't even care. But it killed some time.

vince

Video camera sensor sizes given in specs are left overs from the days of tube cameras. The actual image area is smaller than the full 1/2" (or 2/3").

Here's an article on the subject:

http://www.filmalley.com/articles/sensor%20size/

There is a difference: 35mm motion film exposes the frames transverse to the film strip direction (Academy-Standard) and a 35mm still camera vertical. So the still 35mm pictures have ~ 2.5 x the size of the frames of a 35mm motion camera and therefore less DOF.
So the real difference is approx 1 - 2 stops smaller I guess.