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Author: Subject: Need machine vision camera
Quince
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[*] posted on 10-3-2006 at 22:22


Yes, I can just use hdrshop to get the response curve from multiple exposures. Thanks for your help.
By the way, by RAW you do mean the non-demosaiced Bayer filter pattern image, right?

The one remaining problem is transfer speed. These cameras are USB1.1, so that means quite a few seconds to transfer just one RAW image. It seems only very new cameras use USB2, and only very high end ones use Firewire.

[Edited on 11-3-2006 by Quince]




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[*] posted on 11-3-2006 at 01:41


Quote:
Originally posted by Quince
Yes, I can just use hdrshop to get the response curve from multiple exposures. Thanks for your help.
By the way, by RAW you do mean the non-demosaiced Bayer filter pattern image, right?

Yes, that's what I mean by RAW. The RAW file produced by each camera or family of cameras is closely tied to the hardware, and its format is often not completely documented in an open fashion by the camera manufacturer. That's part of why there are so many different programs, supporting different kinds of cameras, to convert RAW to something more standard like 16 bit TIFF. Existing converters offer quite a bit of flexibility so I would think they can do everything you need.

Edit: It looks like there is already free code that will handle quite a few RAW image types, including those from the G3, if you insist on tinkering with the nuts and bolts.

Quote:
The one remaining problem is transfer speed. These cameras are USB1.1, so that means quite a few seconds to transfer just one RAW image. It seems only very new cameras use USB2, and only very high end ones use Firewire.

That is a drawback. Apparently G3 RAW files are about 3.5 MB, so that's not too long if the camera can sustain near-maximum USB 1.1 transfer rates. It also might be possible to shoot faster by storing images to flash memory and retrieving them only after you have finished the series.

[Edited on 3-11-2006 by Polverone]




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Quince
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[*] posted on 21-3-2006 at 02:57


The problem is I can't use the camera's internal downsampling method as it's far from optimal plus I doubt it can output RAW (unless it has some per-channel binning mode).
How long would it take to transfer one such image? A couple of seconds is OK, more is not.

This sucks, as I'll probably end up waiting till I have a job and spending the thousand bucks for a Prosilica camera.

[Edited on 21-3-2006 by Quince]




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[*] posted on 21-3-2006 at 12:52


Quote:
Originally posted by Quince
The problem is I can't use the camera's internal downsampling method as it's far from optimal plus I doubt it can output RAW (unless it has some per-channel binning mode).

I don't understand what you're saying. The G3 and those other Canons I've listed do output RAW. Downsampling or other transformations of the raw information will take place on your PC.

Quote:
How long would it take to transfer one such image? A couple of seconds is OK, more is not.

This sucks, as I'll probably end up waiting till I have a job and spending the thousand bucks for a Prosilica camera.

A 3.5 MB image would probably take more like 4 seconds to transfer. That's why I suggested the memory card alternative. Put a flash memory card in the camera and remotely trigger it to take pictures saved to the memory card, so you don't have to face the USB bottleneck. When you're done with an item and time is no longer crucial, retrieve the saved images off of the memory card.

Or if you think it's simpler, just wait until you can spend a lot more on a camera you're more confident in, as it seems you may do.




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[*] posted on 21-3-2006 at 14:56


Quote:
I don't understand what you're saying.

What I meant was transferring a lower resolution image (= faster). Some machine vision cameras have binning modes where they bin several pixels into one, sort of emulating a lower resolution but more light sensitive sensor.




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[*] posted on 22-3-2006 at 16:16


USB 1.1 runs at 12MB/s in full speed mode. So 3.5 MB should only take around 1/3 second as a bulk transfer.



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[*] posted on 22-3-2006 at 17:26


That doesn't sound right. It's 12 Mb/s not MB/s, so you're off by a factor of 8. That would make it just over 2 s, but again at full speed. But calculation is one thing, and I'm wondering if it approaches that.

[Edited on 23-3-2006 by Quince]




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[*] posted on 22-3-2006 at 21:22


That'll teach me to listen to my workmate and not look it up. If your camera is on a root hub, with nothing else attached, you should be able to achieve close to that speed. Going though multiple hubs will impact on speed.



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[*] posted on 17-5-2006 at 08:08


While I was in the hospital, the radiologist said for the chest X-rays they used 115 kV. But is that accelerating potential or actual photon energy? And does anyone know how to approximate one from the other for any value? I realize it would vary for different tubes, but I'm wondering if they are similar.



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[*] posted on 17-5-2006 at 20:22


Accelerating voltage. Peak spectral line is probably around the 60kev region. There is no corralation for voltage, only that accelerating potential must be greater. Everything else is mainly intensity and some bremsstrahlung.
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[*] posted on 17-5-2006 at 23:22


The tube I have has a 1 mm aluminum plate as filter. That's very inefficient. Anyone know anything about pulse shaping, so I can have as uniform as possible accelerating potential to monochromatize the emitted spectrum? I've seen designs for pulsed supplies that use large lengths of coaxial cable as blumlein lines for pulse shaping, but I have no clue how to calculate line length and so on. As for doing the actual switching, I have taken out of an X-ray machine (not by me) a pair of high power heatsinked triacs potted in ceramic (AEG T130 N 1200 BOC-17X2, I found no data on these). Anyone have an idea how to drive them?

Marvin, the bremsstrahlung depends on the accelerating potential, and most of the transmitted energy (area under the curve) I see in graphs is under that, rather than characteristic emission, because the characteristics peaks are fairly narrow. So if accelerating potential is increased, I understand the characteristic peaks will remain at the same photon energy and their intensity increase, but the bremsstrahlung should produce higher photon energy. With the right filter applied to the output, all the characteristic peaks could be blocked and only bremsstrahlung can get through. I mean how else to make very hard X-rays, or do they just use different anode materials?

Is there any information somewhere on how emission is proportioned between bremsstrahlung and characteristic peaks depending on configuration?

Without a fairly monochromatic output, tomography is impossible due to differential absorption of various photon energies, and the resulting image will not properly represent densities (unless the sensor is responsive only to a narrow band of X-rays, but I just have a plain scintillation screen).

[Edited on 18-5-2006 by Quince]




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[*] posted on 18-5-2006 at 18:19


The velocity factor for co-ax is typically around 1ns per foot, so you need roughly 1 foot of coax per ns of pulse width.
The transmission line approach can be mimicked by an array of L-C networks, if you don't wish to use 100's of feet of coax. You might also look at "magnetic pulse compression" using saturable inductors for pulse forming.

Are you sure those devices are triacs? I'd have thought thyratrons would be more likely.

Edit: couldn't resist trying a simulation of the L-C network - results in attached pdf :D

[Edited on 19-5-2006 by Twospoons]

Attachment: PULSE FORMING.pdf (203kB)
This file has been downloaded 616 times





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[*] posted on 18-5-2006 at 20:27


One thing is that I have to have the pulse on the primary side of the transformer, simply because the transformer I have is rated for pulse use only and cannot handle continuous mains voltage (regardless of whether there is a load on the secondary). I'm wondering how to compensate for the distortion the transformer makes.

Another concerin is the HV rectifiers. I have ancient GE Kenotrons, though in good condition, I've not yet found what the cathode voltage/current is, so I can build filament transformers for them. Someone suggested to use mechanical synchronous rotary rectifier, but that sounds pretty difficult to build (check this out http://home.freeuk.net/dunckx/wireless/bridge/bridge.html).




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[*] posted on 18-5-2006 at 21:31


Would you mind posting a schematic of what you want to build? It would make providing suggestions easier.

What voltage do you need to rectify?

The pulse forming network should work fine on the primary side of your tranny. I would not expect much distortion, so long as the transformer isn't driven into saturation.




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[*] posted on 18-5-2006 at 21:39


There.

tomo.png - 4kB




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[*] posted on 18-5-2006 at 23:13


Sorry, I wasn't clear - I meant a schematic of your x-ray power supply.



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[*] posted on 19-5-2006 at 00:17


Well I haven't built it, that's why I'm asking here.



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[*] posted on 20-5-2006 at 05:16


Anyone know where I can get a used anti-scatter grid (that's placed between object and shield to remove photons scattered by the object)? Cheapest new grids I found are above $100.



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[*] posted on 13-6-2006 at 11:01


And *this* is why you're not allowed to use the new microwave.

Quote:
Originally posted by Quince
100 keV has less absorption in the body. Much of it will go through.
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[*] posted on 13-6-2006 at 13:16


How long can you wait?

Digital SLRs with decent RAW output and USB 2.0 under $500 (body only though, do you need a lense?) might very well be reality after photokina (september) this year.

Sony stepped into the market which is going to put the prices under pressure.

[Edited on 13-6-2006 by vulture]




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[*] posted on 5-7-2006 at 04:37


Hey guys, I took off the grid from an old computer monitor's picture tube. Is this suitable as an anti-scatter grid, or is the pitch too fine or the depth insufficient?



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[*] posted on 18-5-2007 at 08:12


Hey guys, what determines the resolution of the X-ray system? On the tube I have it says foci 0.8 mm and 1.8 mm. So would that be the size of the smallest feature that can be resolved? I understand that the projected focal spot is smaller than the actual focal spot, but would the numbers on the tube represent the actual or the projected one?
http://compepid.tuskegee.edu/syllabi/clinical/small/radiolog...

[Edited on 18-5-2007 by Nixie]




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