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Very confused on cropping!! Related to HS10

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Kim Letkeman said:
Robsphoto said:
So, using my template, for the arithmetic to “reconcile” properly, the crop factor would be regarded as about 5.84 (36mm divided by 6.16mm).

However, when analysing “real world” examples, there are always “roundings” to contend with, and I guess this is why you say the HS10 has a crop factor of 5.62x?
Click to expand...
And to add further confusion, the crop factor is most useful when it reflects the actual area of the sensor that is used. That way, multiplying the focal length by the crop factor gives an accurate effective focal length.

In this case, Fuji report an actual focal length of 126mm and an effective focal length of 720mm.

That gives a crop factor of 5.7 ... splitting the difference between both of your results.
Click to expand...
The HS10 is quoted as using a 4.2mm – 126mm lens, which in 35mm terms is said to provide a focal length range of 24mm – 720mm.

Therefore, the arithmetical relationship between 4.2mm and 24mm (and 126mm and 720mm) would suggest that the crop factor is 5.714.

But, because the sensor width of the Canon 5D is 36mm, and the HS10 sensor width is quoted as 6.16mm, the arithmetical relationship between these would indicate that the crop factor is 5.844. Therefore, the reach of a 126mm lens on the HS10 may in fact be 736mm, not 720mm. If it’s not 736mm, why would this be the case?

Now, there has been some discussion on this thread about what the dimensions of a 5D image are, after it has been cropped to the same field of view as an image from the HS10. To provide this answer, I will use the format set out in this “template”:

http://www.robsphotography.co.nz/crop-factor-advantage-appendix-3.html

The width of a 5D image is 4368 pixels, so if you divide this by the “real” crop factor of 5.844, you get about 747 pixels (or 4368 x 6.16 / 36 = 747.4).

Then, if you divide this cropped image width of 747 pixels by 1.3333, you get an image height of 560 pixels. (I have divided by 1.3333 because this is the ratio between the width and height of an HS10 image).

Therefore, the cropped image area of a 5D image that has been cropped to the same field of view as an HS 10 image, is only 0.418 megapixels (747 pixels x 560 pixels).

Because the uncropped area of an HS10 image is quoted as 10.3 megapixels, this means that it is about 25 times greater than the cropped 5D image.

It’s no coincidence that the pixel density of the HS10 is also about 25 times greater than that of the 5D (36.19 mp per sq. cm vs 1.47 mp per sq. cm). And, of course, the area of one pixel of the 5D (in square microns) is about 25 times larger than the area of one pixel of the HS10 (68.03 / 2.76).

It’s interesting to note that the image dimensions of the HS10 are quoted as 3648 pixels x 2736 pixels, which you would think represents 9.98 megapixels (3648 x 2736). But the specifications for the HS10 say the HS10 has 10.3 million effective pixels. Can anyone explain how the 10.3 million figure would be calculated?

Regards
Rob
http://www.robsphotography.co.nz/Sony-A900.html
Examples of the amazing resolution of cropped A900 images
 
Robsphoto said:
The HS10 is quoted as using a 4.2mm – 126mm lens, which in 35mm terms is said to provide a focal length range of 24mm – 720mm.

Therefore, the arithmetical relationship between 4.2mm and 24mm (and 126mm and 720mm) would suggest that the crop factor is 5.714.

But, because the sensor width of the Canon 5D is 36mm, and the HS10 sensor width is quoted as 6.16mm, the arithmetical relationship between these would indicate that the crop factor is 5.844. Therefore, the reach of a 126mm lens on the HS10 may in fact be 736mm, not 720mm. If it’s not 736mm, why would this be the case?
Click to expand...
Well, the Cason sensor has a 3:2 aspect ratio and the Fuji sensor has a 4:3 aspect ratio ... there are all sorts of fun possibilities in that difference.

For example ... does the longest side provide the best representation of the crop factor? Or does the diagonal?

The template uses the longest side, but this does not account for the change in image circle size when aspect ratio changes. Perhaps that is why nothing quite adds up?
Robsphoto said:
Now, there has been some discussion on this thread about what the dimensions of a 5D image are, after it has been cropped to the same field of view as an image from the HS10. To provide this answer, I will use the format set out in this “template”:

http://www.robsphotography.co.nz/crop-factor-advantage-appendix-3.html

The width of a 5D image is 4368 pixels, so if you divide this by the “real” crop factor of 5.844, you get about 747 pixels (or 4368 x 6.16 / 36 = 747.4).
Click to expand...
What "real" crop factor? Fuji says 126==720 ... they have spoken. 5.7x (Remember that 3:4 diagonal ... it's a wee bit longer than the longest side, making the ratio ever so slightly smaller)
Robsphoto said:
Then, if you divide this cropped image width of 747 pixels by 1.3333, you get an image height of 560 pixels. (I have divided by 1.3333 because this is the ratio between the width and height of an HS10 image).

Therefore, the cropped image area of a 5D image that has been cropped to the same field of view as an HS 10 image, is only 0.418 megapixels (747 pixels x 560 pixels).

Because the uncropped area of an HS10 image is quoted as 10.3 megapixels, this means that it is about 25 times greater than the cropped 5D image.
Click to expand...
Hmmm ... this all seems rather interesting ... now wht would we want to compare an HS10 at 126mm (720mm effective) against a 5D with a 126mm (126mm effective) lens?

This is not something anyone would do. Ever.

Well ... I did it on an APS-C cam, actually. :-)

http://kimletkeman.blogspot.com/2010/04/hs10-vs-cropped-dslr-megazoom-how-close.html
Robsphoto said:
It’s no coincidence that the pixel density of the HS10 is also about 25 times greater than that of the 5D (36.19 mp per sq. cm vs 1.47 mp per sq. cm). And, of course, the area of one pixel of the 5D (in square microns) is about 25 times larger than the area of one pixel of the HS10 (68.03 / 2.76).
Click to expand...
No ... but what is the point? No one would compare two lenses that have a 6x difference in effective focal length. What is the point?
Robsphoto said:
It’s interesting to note that the image dimensions of the HS10 are quoted as 3648 pixels x 2736 pixels, which you would think represents 9.98 megapixels (3648 x 2736). But the specifications for the HS10 say the HS10 has 10.3 million effective pixels. Can anyone explain how the 10.3 million figure would be calculated?
Click to expand...
The Sony WX1 uses the same sensor and its spec sheet shows 10.6m sensor pixels, 10.2m effective pixels, and the same maximum size of 9.98mp.

My guess is that we have a sensor that uses some pixels around the side for various processing such as digital stabilization or what have you. It may also have something to so with how they process different aspect ratios.

--
http://kimletkeman.blogspot.com
http://letkeman.net/Photos
 
Robsphoto said:
Kim Letkeman said:
Robsphoto said:
So, using my template, for the arithmetic to “reconcile” properly, the crop factor would be regarded as about 5.84 (36mm divided by 6.16mm).

However, when analysing “real world” examples, there are always “roundings” to contend with, and I guess this is why you say the HS10 has a crop factor of 5.62x?
Click to expand...
And to add further confusion, the crop factor is most useful when it reflects the actual area of the sensor that is used. That way, multiplying the focal length by the crop factor gives an accurate effective focal length.

In this case, Fuji report an actual focal length of 126mm and an effective focal length of 720mm.

That gives a crop factor of 5.7 ... splitting the difference between both of your results.
Click to expand...
The HS10 is quoted as using a 4.2mm – 126mm lens, which in 35mm terms is said to provide a focal length range of 24mm – 720mm.

Therefore, the arithmetical relationship between 4.2mm and 24mm (and 126mm and 720mm) would suggest that the crop factor is 5.714.

But, because the sensor width of the Canon 5D is 36mm, and the HS10 sensor width is quoted as 6.16mm, the arithmetical relationship between these would indicate that the crop factor is 5.844. Therefore, the reach of a 126mm lens on the HS10 may in fact be 736mm, not 720mm. If it’s not 736mm, why would this be the case?
Click to expand...
Full-Frame is 36mm while the Canon 5D is close, it is not exact (35.8mm). Also you should not use the sensor width for accurate crop factoring and use diagonal measure instead. And in case of sensors with 4:3 aspect ratio, you must also use the average diagonal measure of the 4:3 and 3:2 as reference.
Robsphoto said:
Now, there has been some discussion on this thread about what the dimensions of a 5D image are, after it has been cropped to the same field of view as an image from the HS10. To provide this answer, I will use the format set out in this “template”:

http://www.robsphotography.co.nz/crop-factor-advantage-appendix-3.html

The width of a 5D image is 4368 pixels, so if you divide this by the “real” crop factor of 5.844, you get about 747 pixels (or 4368 x 6.16 / 36 = 747.4).
Click to expand...
You should go by the calculated crop factor stated on the lens which is 5.7(14) as the "real" crop factor.
Robsphoto said:
Then, if you divide this cropped image width of 747 pixels by 1.3333, you get an image height of 560 pixels. (I have divided by 1.3333 because this is the ratio between the width and height of an HS10 image).

Therefore, the cropped image area of a 5D image that has been cropped to the same field of view as an HS 10 image, is only 0.418 megapixels (747 pixels x 560 pixels).

Because the uncropped area of an HS10 image is quoted as 10.3 megapixels, this means that it is about 25 times greater than the cropped 5D image.

It’s no coincidence that the pixel density of the HS10 is also about 25 times greater than that of the 5D (36.19 mp per sq. cm vs 1.47 mp per sq. cm). And, of course, the area of one pixel of the 5D (in square microns) is about 25 times larger than the area of one pixel of the HS10 (68.03 / 2.76).
Click to expand...
Of course, they are all relative.
Robsphoto said:
It’s interesting to note that the image dimensions of the HS10 are quoted as 3648 pixels x 2736 pixels, which you would think represents 9.98 megapixels (3648 x 2736). But the specifications for the HS10 say the HS10 has 10.3 million effective pixels. Can anyone explain how the 10.3 million figure would be calculated?
Click to expand...
9.98 is right for a 10MP with 4:3 aspect ratio. Those extra pixels are needed for cushion and other sensor functions including the needed space to compensate for lens distortion.

...
 
Thanks very much Kim and RoysLaw for your comments on my recent post.

On the authoritative web site linked to below, it states that the crop factor refers to the size of the sensor in a digital camera relative to the size of the frame in a 35mm camera, which has a frame size of 36mm x 24mm.

http://www.clarkvision.com/photoinfo/cropfactor/

Sometimes, before dealing with “real world” examples, it can be helpful to look at a simpler “theoretical” model. So, here’s a simple theoretical example, which features cameras that have similar specifications to the HS10 and the 5D.

So, let's assume that the smaller camera has a 4:3 aspect ratio, and is referred to as “SS”.

The larger full frame camera has a 3:2 aspect ratio and is referred to as “FF”.

It is assumed that both cameras have a 100mm lens attached, and that the longer sides of the images quoted below are regarded as the “widths” of the images.

Assume that SS has a sensor size of 6mm x 4.5mm, image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Assume that FF has a sensor size of 36mm x 24mm, image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

We know that the field of view of FF when it has a 100mm lens attached is 100mm, because there is no crop factor as it has a sensor size of 36mm x 24mm.

So, the questions we are considering are:

1. What is the field of view (in mm) of SS when it has a 100mm lens attached, and how is this calculated?”

2. If an image from FF is cropped to the same field of view as an image from SS, what are its revised dimensions?

I would be interested in your views with regard to these questions.

Thanks very much.

Regards
Rob
http://www.robsphotography.co.nz
 
Robsphoto said:
On the authoritative web site linked to below, it states that the crop factor refers to the size of the sensor in a digital camera relative to the size of the frame in a 35mm camera, which has a frame size of 36mm x 24mm.

http://www.clarkvision.com/photoinfo/cropfactor/
Click to expand...
Clarkvision is a great site. And yes, the crop factor is defined that way. But when a manufacturer tells you the conversion from their focal lengths to the 35mm equivalents, they are telling you the crop factor as well. So minor differences between that and what you see in the size of the wafer can be chocked up to unknown causes ... the engineers could tell you, but they won't.
Robsphoto said:
Sometimes, before dealing with “real world” examples, it can be helpful to look at a simpler “theoretical” model. So, here’s a simple theoretical example, which features cameras that have similar specifications to the HS10 and the 5D.

So, let's assume that the smaller camera has a 4:3 aspect ratio, and is referred to as “SS”.

The larger full frame camera has a 3:2 aspect ratio and is referred to as “FF”.

It is assumed that both cameras have a 100mm lens attached, and that the longer sides of the images quoted below are regarded as the “widths” of the images.

Assume that SS has a sensor size of 6mm x 4.5mm, image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Assume that FF has a sensor size of 36mm x 24mm, image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

We know that the field of view of FF when it has a 100mm lens attached is 100mm, because there is no crop factor as it has a sensor size of 36mm x 24mm.

So, the questions we are considering are:

1. What is the field of view (in mm) of SS when it has a 100mm lens attached, and how is this calculated?”
Click to expand...
I'm rather sure that I asked you that question :-)

Do you look only at the longest side?

Or is the diagonal more accurate, since it factors in the difference in aspect ration?

I'm going with the diagonal if that's all you got. If aspect ratios are the same, then the linear dimensions work fine, since the ratios are the same all around.
Robsphoto said:
2. If an image from FF is cropped to the same field of view as an image from SS, what are its revised dimensions?
Click to expand...
Depends on what you come up with for the crop factor. Either way, each dimension is now 1/CF of the original I believe.

--
http://kimletkeman.blogspot.com
http://letkeman.net/Photos
 
Mister T said:
DrewE said:
Mister T said:
... and as I might have mentioned before, doesn't a 10mp sensor have 2.5m red cells, 2.5m blue cells and 5m green cells so really that's only 2.5m pixels? So up-scaling is a well practised and mature science that fools us all?
Click to expand...
Pixels are primarily defined spatially, not in terms of color capability. I have an old Macintosh with a 512x384 screen where each pixel can be either black or white; yet they are still pixels.

A 10MP bayer sensor does indeed have a photosite breakdown by color as you outline, but it does capture light at 10M spatially distinct points, and so it's not a stretch to call it 10 megapixels. It cannot resolve color differences at the full resolution, but can (nearly) resolve luminance differences at full resolution, and the eye tends to be more sensitive to luminance details than chromatic details. (Many JPEG files store color information at a lower resolution than luminance information for this very reason.)

With a Bayer sensor, it is of course necessary to interpolate the color information (called "demosaicing") in producing a usable image. This isn't the same as upscaling, since there is some image data to guide the process.
Click to expand...
Point taken, should have noticed this. This is much the same way that colour information is put in the PAL TV signal (I don't know if NTSC is the same) where the 6MHz bandwidth, originally designed just for luminance in black and white days, has the colour signal 'hidden' in it but at a lower resolution.

Is it fair to say that the colour resolution is lower and that areas of similar brightness with fine colour patterns will suffer a bit of 'mushiness' (good technical term there)?
Click to expand...
That's absolutely fair and correct, assuming a well-behaved demosaicing algorithm. Similarly, areas of very fine luminance detail can cause false color artifacts. Luckily, in practice, both tend to be the exception rather than the rule. On many modern compact cameras, it's virtually impossible to get either one because the lens is incapable of resolving such fine details (due to diffraction limitations, if nothing else).
--
--DrewE
 
Robsphoto said:
Thanks very much Kim and RoysLaw for your comments on my recent post.

On the authoritative web site linked to below, it states that the crop factor refers to the size of the sensor in a digital camera relative to the size of the frame in a 35mm camera, which has a frame size of 36mm x 24mm.

http://www.clarkvision.com/photoinfo/cropfactor/

Sometimes, before dealing with “real world” examples, it can be helpful to look at a simpler “theoretical” model. So, here’s a simple theoretical example, which features cameras that have similar specifications to the HS10 and the 5D.

So, let's assume that the smaller camera has a 4:3 aspect ratio, and is referred to as “SS”.

The larger full frame camera has a 3:2 aspect ratio and is referred to as “FF”.

It is assumed that both cameras have a 100mm lens attached, and that the longer sides of the images quoted below are regarded as the “widths” of the images.

Assume that SS has a sensor size of 6mm x 4.5mm, image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Assume that FF has a sensor size of 36mm x 24mm, image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

We know that the field of view of FF when it has a 100mm lens attached is 100mm, because there is no crop factor as it has a sensor size of 36mm x 24mm.

So, the questions we are considering are:

1. What is the field of view (in mm) of SS when it has a 100mm lens attached, and how is this calculated?”

2. If an image from FF is cropped to the same field of view as an image from SS, what are its revised dimensions?

I would be interested in your views with regard to these questions.

Thanks very much.

Regards
Rob
http://www.robsphotography.co.nz
Click to expand...
If you apply approximation or round-of to one decimal, they would all give you the same result wether you measure by width, length or diagonal.

But to answer your question, it depends on the camera design and how the image circle is applied. For fixed lens cameras with 4:3 format sensor, how the image circle is applied is usually unknown. Meaning, is the 3:2 image a crop of 4:3 image or is it the other way around? In marketing driven, "real world" answer, I would bet that in most cases, what is stated on the lens is the one that yields the largest number.

...
 
RoysLaw said:
Robsphoto said:
Thanks very much Kim and RoysLaw for your comments on my recent post.

On the authoritative web site linked to below, it states that the crop factor refers to the size of the sensor in a digital camera relative to the size of the frame in a 35mm camera, which has a frame size of 36mm x 24mm.

http://www.clarkvision.com/photoinfo/cropfactor/

Sometimes, before dealing with “real world” examples, it can be helpful to look at a simpler “theoretical” model. So, here’s a simple theoretical example, which features cameras that have similar specifications to the HS10 and the 5D.

So, let's assume that the smaller camera has a 4:3 aspect ratio, and is referred to as “SS”.

The larger full frame camera has a 3:2 aspect ratio and is referred to as “FF”.

It is assumed that both cameras have a 100mm lens attached, and that the longer sides of the images quoted below are regarded as the “widths” of the images.

Assume that SS has a sensor size of 6mm x 4.5mm, image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Assume that FF has a sensor size of 36mm x 24mm, image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

We know that the field of view of FF when it has a 100mm lens attached is 100mm, because there is no crop factor as it has a sensor size of 36mm x 24mm.

So, the questions we are considering are:

1. What is the field of view (in mm) of SS when it has a 100mm lens attached, and how is this calculated?”

2. If an image from FF is cropped to the same field of view as an image from SS, what are its revised dimensions?

I would be interested in your views with regard to these questions.

Thanks very much.

Regards
Rob
http://www.robsphotography.co.nz
Click to expand...
If you apply approximation or round-of to one decimal, they would all give you the same result wether you measure by width, length or diagonal.

But to answer your question, it depends on the camera design and how the image circle is applied. For fixed lens cameras with 4:3 format sensor, how the image circle is applied is usually unknown. Meaning, is the 3:2 image a crop of 4:3 image or is it the other way around? In marketing driven, "real world" answer, I would bet that in most cases, what is stated on the lens is the one that yields the largest number.

...
Click to expand...
Thanks very much for your reply. For what it's worth, here is the approach that I would take to solve the questions I asked with regard to the above theoretical example.

When you compare the width of the sensor of FF (36mm) with that of SS (6mm), the crop factor is 6.

But, if you compare the height of the sensor of FF (24mm) with that of SS (4.5mm), the crop factor is 5.3333.

When making the various LINEAR calculations, I would use the crop factor of 6 (rather an "average" or "diagonal" crop factor). When a 100mm lens is used on SS, this would provide a field of view (FOV) of 600mm.

The width of an image of FF that has been cropped to the same FOV as an image of SS, is 800 pixels (4800 divided by the “linear” crop factor of 6).

The height of an image of FF that has been cropped to the same FOV as an image of SS, is 600 pixels. This represents the original height of 3200 pixels divided by the “height” crop factor of 5.3333 (or 800 pixels width / 1.3333).

If the above approach is adopted, all the various linear and area relationships are consistently maintained, as set out in my template here:

http://www.robsphotography.co.nz/crop-factor-advantage-appendix-3.html

However, I realise that there can be different interpretations on how to calculate the crop factor when a camera with a 4:3 aspect ratio is compared with a full frame camera which has a 3:2 aspect ratio!

Regards
Rob
 
Ok ... you've repeated the calculations you like and the pointer to the template 3 times in this thread. I think we all get it now.

But your template still comes up with the wrong number for the Fuji HS10, at least according to Fuji.

So I think hard and fast rules don;t really work here ... it's all approximations.

--
http://kimletkeman.blogspot.com
http://letkeman.net/Photos
 
Kim Letkeman said:
Robsphoto said:
On the authoritative web site linked to below, it states that the crop factor refers to the size of the sensor in a digital camera relative to the size of the frame in a 35mm camera, which has a frame size of 36mm x 24mm.

http://www.clarkvision.com/photoinfo/cropfactor/
Click to expand...
Clarkvision is a great site. And yes, the crop factor is defined that way. But when a manufacturer tells you the conversion from their focal lengths to the 35mm equivalents, they are telling you the crop factor as well. So minor differences between that and what you see in the size of the wafer can be chocked up to unknown causes ... the engineers could tell you, but they won't.
Kim Letkeman said:
Sometimes, before dealing with “real world” examples, it can be helpful to look at a simpler “theoretical” model. So, here’s a simple theoretical example, which features cameras that have similar specifications to the HS10 and the 5D.

So, let's assume that the smaller camera has a 4:3 aspect ratio, and is referred to as “SS”.

The larger full frame camera has a 3:2 aspect ratio and is referred to as “FF”.

It is assumed that both cameras have a 100mm lens attached, and that the longer sides of the images quoted below are regarded as the “widths” of the images.

Assume that SS has a sensor size of 6mm x 4.5mm, image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Assume that FF has a sensor size of 36mm x 24mm, image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

We know that the field of view of FF when it has a 100mm lens attached is 100mm, because there is no crop factor as it has a sensor size of 36mm x 24mm.

So, the questions we are considering are:

1. What is the field of view (in mm) of SS when it has a 100mm lens attached, and how is this calculated?”
Click to expand...
I'm rather sure that I asked you that question :-)

Do you look only at the longest side?

Or is the diagonal more accurate, since it factors in the difference in aspect ration?

I'm going with the diagonal if that's all you got. If aspect ratios are the same, then the linear dimensions work fine, since the ratios are the same all around.
Kim Letkeman said:
2. If an image from FF is cropped to the same field of view as an image from SS, what are its revised dimensions?
Click to expand...
Depends on what you come up with for the crop factor. Either way, each dimension is now 1/CF of the original I believe.

--
http://kimletkeman.blogspot.com
http://letkeman.net/Photos
Click to expand...
Thanks very much for your reply. If you use a diagonal measurement, then you would obtain a crop factor of about 5.62 for the Fujifilm HS10 (43.267mm / 7.7mm as explained below).

The HS10 has a sensor size of 6.16mm x 4.62mm and a diagonal measurement of 7.7mm. In contrast, a full frame camera has a sensor size of 36mm x 24mm and a diagonal measurement of 43.267mm.

If you use a 126mm focal length on the HS10, this would provide a field of view (FOV) of 708mm (126 x 5.62).

But, the HS10 is quoted as having a FOV of 720mm with a 126mm lens, which gives a crop factor of 5.714.

If you calculate the crop factor of the HS10 by dividing the width of a full frame sensor (36mm) by the width of the HS10 sensor (6.16mm), then the crop factor is 5.844. If you use a 126mm focal length on the HS10, this would provide a FOV of 736mm. I favour this approach for the reasons explained here:

http://forums.dpreview.com/forums/read.asp?forum=1012&message=35357059

You could say that the difference between FOVs of 708mm, 720mm, and 736mm is quite small and only of academic interest!

Regards
Rob
http://www.robsphotography.co.nz/crop-factor-advantage.html
Article about the telephoto advantage of an APS-C camera
 
Robsphoto said:
Thanks very much for your reply. If you use a diagonal measurement, then you would obtain a crop factor of about 5.62 for the Fujifilm HS10 (43.267mm / 7.7mm as explained below).

The HS10 has a sensor size of 6.16mm x 4.62mm and a diagonal measurement of 7.7mm. In contrast, a full frame camera has a sensor size of 36mm x 24mm and a diagonal measurement of 43.267mm.

If you use a 126mm focal length on the HS10, this would provide a field of view (FOV) of 708mm (126 x 5.62).

But, the HS10 is quoted as having a FOV of 720mm with a 126mm lens, which gives a crop factor of 5.714.

If you calculate the crop factor of the HS10 by dividing the width of a full frame sensor (36mm) by the width of the HS10 sensor (6.16mm), then the crop factor is 5.844. If you use a 126mm focal length on the HS10, this would provide a FOV of 736mm. I favour this approach for the reasons explained here:

http://forums.dpreview.com/forums/read.asp?forum=1012&message=35357059

You could say that the difference between FOVs of 708mm, 720mm, and 736mm is quite small and only of academic interest!

Regards
Rob
http://www.robsphotography.co.nz/crop-factor-advantage.html
Article about the telephoto advantage of an APS-C camera
Click to expand...
Like I said, it is hard to be very specific with calculations without knowing how things are applied. But being a number person myself, it sure bugs me when things just don't add up to the T, so I feel for your frustration.

So let's work with what's given and this will explain exactly what's going on and how the calc is applied by Fuji engineers.

This will be a "short" version but since you are a math guy I'm sure you can follow:
  • HS10 sensor size is 6.16mm x 4.62mm @ 10.3MP
  • Recorded pixel size is 9.98MP
---- So to calculate: 10.3/9.98 = 1.032 and the square root of that = 1.016
  • Recalculate sensor: 6.16/1.016 = 6.06 and 4.62/1.016 = 4.55
  • Diagonal measure for 6.06x4.54 = 7.58
  • 43.27/7.58 = 5.71
So there you go. I hope that clarifies that now. =)

...
 
RoysLaw said:
Robsphoto said:
Thanks very much for your reply. If you use a diagonal measurement, then you would obtain a crop factor of about 5.62 for the Fujifilm HS10 (43.267mm / 7.7mm as explained below).

The HS10 has a sensor size of 6.16mm x 4.62mm and a diagonal measurement of 7.7mm. In contrast, a full frame camera has a sensor size of 36mm x 24mm and a diagonal measurement of 43.267mm.

If you use a 126mm focal length on the HS10, this would provide a field of view (FOV) of 708mm (126 x 5.62).

But, the HS10 is quoted as having a FOV of 720mm with a 126mm lens, which gives a crop factor of 5.714.

If you calculate the crop factor of the HS10 by dividing the width of a full frame sensor (36mm) by the width of the HS10 sensor (6.16mm), then the crop factor is 5.844. If you use a 126mm focal length on the HS10, this would provide a FOV of 736mm. I favour this approach for the reasons explained here:

http://forums.dpreview.com/forums/read.asp?forum=1012&message=35357059

You could say that the difference between FOVs of 708mm, 720mm, and 736mm is quite small and only of academic interest!

Regards
Rob
http://www.robsphotography.co.nz/crop-factor-advantage.html
Article about the telephoto advantage of an APS-C camera
Click to expand...
Like I said, it is hard to be very specific with calculations without knowing how things are applied. But being a number person myself, it sure bugs me when things just don't add up to the T, so I feel for your frustration.

So let's work with what's given and this will explain exactly what's going on and how the calc is applied by Fuji engineers.

This will be a "short" version but since you are a math guy I'm sure you can follow:
  • HS10 sensor size is 6.16mm x 4.62mm @ 10.3MP
  • Recorded pixel size is 9.98MP
---- So to calculate: 10.3/9.98 = 1.032 and the square root of that = 1.016
  • Recalculate sensor: 6.16/1.016 = 6.06 and 4.62/1.016 = 4.55
  • Diagonal measure for 6.06x4.54 = 7.58
  • 43.27/7.58 = 5.71
So there you go. I hope that clarifies that now. =)
Click to expand...
Thanks for posting this explanation, it’s very interesting. The focal length range of the lens on the HS10, in 35mm terms is said to be 126mm – 720mm, which represents a crop factor of 5.71 (720 / 126). You say that this crop factor is based on a comparison of the diagonal measure of the sensors of the two cameras (43.27 / 7.58).

I am not sure that a lens “reach” calculation based on a “diagonally measured” crop factor is as accurate as a calculation made with a crop factor that is based on a comparison of the widths of the two sensors (36mm / 6.06mm), but as stated earlier, the difference in the answers is not all that great, and not enough to affect a buying decision!

The formulas that I have used for calculating the effective “reach” of cropped images are shown here:

http://www.robsphotography.co.nz/focal-length.html

Regards
Rob
 
Robsphoto said:
RoysLaw said:
Robsphoto said:
Thanks very much for your reply. If you use a diagonal measurement, then you would obtain a crop factor of about 5.62 for the Fujifilm HS10 (43.267mm / 7.7mm as explained below).

The HS10 has a sensor size of 6.16mm x 4.62mm and a diagonal measurement of 7.7mm. In contrast, a full frame camera has a sensor size of 36mm x 24mm and a diagonal measurement of 43.267mm.

If you use a 126mm focal length on the HS10, this would provide a field of view (FOV) of 708mm (126 x 5.62).

But, the HS10 is quoted as having a FOV of 720mm with a 126mm lens, which gives a crop factor of 5.714.

If you calculate the crop factor of the HS10 by dividing the width of a full frame sensor (36mm) by the width of the HS10 sensor (6.16mm), then the crop factor is 5.844. If you use a 126mm focal length on the HS10, this would provide a FOV of 736mm. I favour this approach for the reasons explained here:

http://forums.dpreview.com/forums/read.asp?forum=1012&message=35357059

You could say that the difference between FOVs of 708mm, 720mm, and 736mm is quite small and only of academic interest!

Regards
Rob
http://www.robsphotography.co.nz/crop-factor-advantage.html
Article about the telephoto advantage of an APS-C camera
Click to expand...
Like I said, it is hard to be very specific with calculations without knowing how things are applied. But being a number person myself, it sure bugs me when things just don't add up to the T, so I feel for your frustration.

So let's work with what's given and this will explain exactly what's going on and how the calc is applied by Fuji engineers.

This will be a "short" version but since you are a math guy I'm sure you can follow:
  • HS10 sensor size is 6.16mm x 4.62mm @ 10.3MP
  • Recorded pixel size is 9.98MP
---- So to calculate: 10.3/9.98 = 1.032 and the square root of that = 1.016
  • Recalculate sensor: 6.16/1.016 = 6.06 and 4.62/1.016 = 4.55
  • Diagonal measure for 6.06x4.54 = 7.58
  • 43.27/7.58 = 5.71
So there you go. I hope that clarifies that now. =)
Click to expand...
Thanks for posting this explanation, it’s very interesting. The focal length range of the lens on the HS10, in 35mm terms is said to be 126mm – 720mm, which represents a crop factor of 5.71 (720 / 126). You say that this crop factor is based on a comparison of the diagonal measure of the sensors of the two cameras (43.27 / 7.58).

I am not sure that a lens “reach” calculation based on a “diagonally measured” crop factor is as accurate as a calculation made with a crop factor that is based on a comparison of the widths of the two sensors (36mm / 6.06mm), but as stated earlier, the difference in the answers is not all that great, and not enough to affect a buying decision!
Click to expand...
Why wouldn't it be? The proof is right there in front of you. Diagonal measure is the average of length and width. If the sensor in reference has the exact aspect ratio then it doesn't matter as all sides are proportional. However not all sensor are built the same.
Robsphoto said:
The formulas that I have used for calculating the effective “reach” of cropped images are shown here:

http://www.robsphotography.co.nz/focal-length.html

Regards
Rob
Click to expand...
 
Robsphoto said:
Thanks for posting this explanation, it’s very interesting. The focal length range of the lens on the HS10, in 35mm terms is said to be 126mm – 720mm, which represents a crop factor of 5.71 (720 / 126). You say that this crop factor is based on a comparison of the diagonal measure of the sensors of the two cameras (43.27 / 7.58).

I am not sure that a lens “reach” calculation based on a “diagonally measured” crop factor is as accurate as a calculation made with a crop factor that is based on a comparison of the widths of the two sensors (36mm / 6.06mm), but as stated earlier, the difference in the answers is not all that great, and not enough to affect a buying decision!
Click to expand...
Forgive me .... but you come across as completely married to your own theory that width is all that matters.

The fact is that the diagonal makes sense to both me and royslaw as it accounts for aspect ratio. The fact that he was able to come up with a calculation using the diagonal that mapped to the actual crop factor as reported by Fuji lends further weight to the diagonal as the better crop factor determinant.

And here is the final word (I hope) , showing that the diagonal is a perfect reflection of the image circles required for two sensors of the same width but with different aspect ratios.

If you still dispute the diagonal after seeing this image, then I will shake my head in wonder ...


Robsphoto said:
The formulas that I have used for calculating the effective “reach” of cropped images are shown here:

http://www.robsphotography.co.nz/focal-length.html

Regards
Rob
Click to expand...
--
http://kimletkeman.blogspot.com
http://letkeman.net/Photos
 
Yep... calculation of crop factor should be made considering diagonal and not horizontal or vertical field.... this allows a more reliable comparison between lens image circle...

That diagram explains it quite well :3

I used diagonal as well for my (confusing i would say) calculation in this same thread but other discussion branch:
http://forums.dpreview.com/forums/read.asp?forum=1012&message=35316576

Note however that cropping digitally maintaining aspect ratio of original will give a crop ratio proportional not only from diagonal but from horizontal and vertical fields as well... funny math stuff happens :D
 
Thanks for your comments, it’s been a most interesting discussion!

I quoted earlier this theoretical example, which features a camera (referred to as SS) that has similar specifications to the HS10.

Camera SS has a sensor size of 6mm x 4.5mm, an image size of 3600 pixels x 2700 pixels, and 9.72 megapixels.

Camera FF has a sensor size of 36mm x 24mm, an image size of 4800 pixels x 3200 pixels, and 15.36 megapixels.

If you calculate the crop factor using diagonal measurements, it is 5.769 (43.267 / 7.5).

I have quickly put up a temporary page on internet that applies the above information on a “template” that I referred to earlier in this thread. This shows that the linear pixel density of SS is 4.5 times greater than that of FF. You can see this temporary page here:

http://www.robsphotography.co.nz/diagonal.html

In the examples I have done with cameras that have a 3:2 aspect ratio, all the linear results are consistent with this 4.5 times difference between the results of SS and FF.

But, when you have a 4:3 aspect ratio, if you use a crop factor of 5.769 (instead of a “linear crop factor” of 6.0), not all the results are consistent with the fact that the linear pixel density of SS is 4.5 times greater than that of FF.

This raises the question of whether a “non linear” crop factor should have been used when making the linear comparisons shown on the above page?

QUESTION: Do you think the linear and area comparisons made on the above temporary web page are satisfactory and in line with what you would expect?

Thanks for your views on this question.

Regards
Rob
 
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