Subject size difference at 300mm & 500mm

<p>Hi, I'm very new to photography so please be gentle. I have a Canon 40D and have been using a Canon 70 - 300mm is lense. I recently photographed an Osprey, which was about 800m away and was disappointed by the size of the bird in the resulting photograph.</p>

<p>My question is would the Tamron 200-500mm used at 500mm make the subject - in this case the Osprey -appear that much bigger to justify the price. Or is there a better alternative or should I just forget it. I am unable to get any closer to the birds. Please help.</p>

<p>Thanks</p>

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<p>http://www.tamron.com/lenses/learning_center/tools/focal-length-comparison.php</p>

<p>You can see the difference between 300 and 500 with this tool for that exact lens.</p>

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<p>The osprey would be 1.6 times as big with the 500mm.</p>

<p>Just a quick calculation...let the wingspan of the osprey be 3meters, at a distance of 800meters, you need a focal length of 800m / 3m * 22mm = 5800mm (!) to have the wingspan fill the frame horizontally. With a 500mm you have the wingspan on less than 10percent of the horizontal frame.</p>

<p>You should get nearer.</p>

 

<p>Linear magnification is proportional to the focal length. At the same distance, the 500mm will give you an image 5/3 wider and 5/3 taller compared to the 300mm.</p>

<p>Thanks Rob, I'll have a look at that.</p>

<p>Thanks Rainer T & Canon user. </p>

<p>Thanks again. Been thinking about it and would a prime lens say the Canon EF 400mm f5.6L USM give me sharp enough shots to crop and enlarge for a decent sized print.</p>

<p>With the 400mm lens, the ospray will probably cover 1/15 of the horizontal size ... which is about 280 pixels wide ... no, I'm afraid, this is not enough to get a print. 800meters distance is simply too far away.</p>

<p >You are asking about magnification: When “birding” we use binoculars or a telephoto to magnify (make the object appear to be closer). As an example a bird 800 meters away will appear as if it is only 100 meters away if observed using a pair of 8x binoculars (800 ÷ 8 = 100). If you used a 20x telescope the object would appear to be only 40 meters distance. </p>

<p > </p>

<p >When a camera is utilized with a telephoto lens the magnification achieved is not as clear cut. The astronomical community traditionally divides the focal length by 50 to derive the magnification. Thus a 300mm lens delivers 300 ÷ 50 = 6 power (6x). Meaning the bird will photograph the same size as if you had closed the distance to 800 ÷ 6 = 130 meters. Using this logic a 500mm delivers 500 ÷ 50 = 10x power thus the bird 800 meters ÷ 10 = 80 meters apparent distance. </p>

<p > </p>

<p >Now the 50 divisor comes from the usual and customary prime lens (50mm) normally mounted on a 35mm film camera. This value is a rounding upward of the diagonal measure of the film frame which measures 24mm by 36mm. The diagonal of this film format measures 43.2mm. This value is an out of the ordinary value that opticians round up to 50mm.</p>

<p > </p>

<p >Now your 40D sports a smaller sensor (format size), smaller than the 35mm film format. Your camera’s format (sensor) measures 14.8mm by 22.2 and the diagonal measure is 26.7mm. Using the same logic a 300mm delivers 300 ÷ 26.7 = 11x or 11 power. The bird would appear to be 800 meters ÷ 11 = 73 meters away. If you mount a 500mm the math is 500 ÷ 26.7 = 19x or 19 power. Thus the 500mm causes the bird to appear as if it were 26 meters away.</p>

<p > </p>

<p >The 40D sensor size is called APS-C (Advanced Photo System Classic Format). Often you will read about a crop factor or magnification factor. It is used to make comparisons as to lens performance, your camera to a full frame 35mm. This value is 43.2 ÷ 26.7 = 1.6. How do you use this value? Given a 300mm it will perform on your camera the same as a 300 x 1.6 = 480mm will on a full frame. Stated another way; the smaller sensor of your 40D yields 1.6 x more magnification than the same lens mounted on a full frame 35mm.</p>

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<p >Once you have taken the picture you will be viewing it on a computer screen or by print. The size of the exhibit must be taken into account. Say you made a 4x6 inch print. This is an enlargement. Actually if un-cropped it is about a 4 times enlargement this is additional power (magnification). Thus the bird shot with a 300 using the 40D and printed 4x6 inches appears to be 73 meters ÷ 4 = 18 meters away. </p>

<p > </p>

<p >Nobody said this stuff is easy! And many will question this math so let me say in advance – photography is both an art and a science. You don’t need to know both but it helps. </p>

<p>Thanks Rainer T - I realise this now. <br>

Thanks also to Alan Marcus for your brilliant explanation. I will copy the text and print it off to re-read and hopefully fully understand and remember.<br>

Thanks again.</p>

 

<p>There is no substitute for getting close.</p>

<p>For these magnifications, do a search on "digiscoping". It is an easier technique plus it can help you lessen equipment weight and cost. However, don't expect miracle. You still can't break the law of optical physic.</p>

<p>Sorry, but you're way too far away. Take a look at some of Doug Herr's bird photos to get a sense of what's possible by using a long lens and stalking to get close for winning shots.</p>

<p>Marjory, I have a couple Osprey images in my portfolio, one in the "Mountains" folder and the other in the "Birds" folder. Both were taken using a Canon 5D and 100-400 at 400mm. You would get some crop by using your 40D but as Kerry suggested, there really is no substitute for getting closer. Both images in my portfolio are heavily cropped.</p>

<p>Many thanks to all who replied. I now know I need to lower my expectations of what is achievable. </p>

Alan, thanks so much for such a thorough explanation. I’ve had similar questions about the difference in focal length magnification. I’ve also found as other have mentioned how important it is to get as close as possible to one’s subject regardless of focal length. Based on the formulas you have stated above, it would appear that given a distance of 50’ (15.2 meters), and shooting with a 300 mm lens, one would only need to move approximately 3’ closer to the subject to obtain the same magnification as using a 500mm at the same distance. Correct?

Here is a useful tool from Nikon - Link Nikon | Imaging Products | NIKKOR Lens Simulator

800m is very far. I don't think the 500mm would make you happy.

<p>Many thanks to all who replied. I now know I need to lower my expectations of what is achievable. </p>

Do NOT lower your expectations of what is achievable. Instead, raise your skills. Mainly, get a lot closer, add patience, and keep at it. Remember that for all those great photos of wildlife that you see, the photographer is only showing you his best.

given a distance of 50’ (15.2 meters), and shooting with a 300 mm lens, one would only need to move approximately 3’ closer to the subject to obtain the same magnification as using a 500mm at the same distance. Correct?

No. 500mm at 50' has the same FOV as 300mm at 30' - you need to get 20' closer to the subject. Focal lengths differ by a factor of 1.6 - and to get equal subject size, the distance needs to change by the same factor.

The camera lens acts like a movie or slide projector in that it projects an image on a screen. In the case of a camera, the screen is film or digital sensor. Unlike the movie projector system which projects a magnified view, the camera lens projects a miniature image of the outside world.

 

Using the math related to triangles, we can easily figure out how big the image of an object will be. To do this we must know how far away the object is and how tall the object is. For birds, we can estimate how far away they are. Suppose the bird is 1 ½ feet tall = 1.5 feet --- and the bird is 300 feet distant. We can draw an imaginary triangle from the bird to the camera lens. This triangle is called the object triangle. The base of this triangle is 1.5 and height of this triangle is 300. Now the ratio base to height is 1.5 ÷ 300 = 0.005. This is a ratio; thus it has no dimensions. However we now know the key number to work the problem is 0.005.

 

Now the lens projects the image of the bird. We can also trace out an imaginary triangle called the image triangle. Using a 300mm lens we can use this value for the height of the image triangle. We can now compute the height of the image of bird. It will be 300 X 0.005 = 1.5mm (not very big, an inch is 25.4mm). If we switch to a 500mm lens the revised image of bird is now 500 X 0.005 = 2.5mm.

 

Suppose we borrow or rent a 1000mm camera lens; now the bird’s image will be 1000 X 0.005 = 5mm.

 

This is just a small example of the gobbledygook I can throw around.

Suppose we borrow or rent a 1000mm camera lens; now the bird’s image will be 1000 X 0.005 = 5mm.

And a lot fuzzier if the 1000mm lens in question is of catadioptric construction!;)

 

Seriously though, I bought a 1000mm f/11 Reflex-Nikkor mirror lens; a) because I was impressed by its size, b) because it was relatively cheap and, c) because I thought it would be useful and deliver better 'close ups'. Assumption c) proved to be utterly wrong.

 

Distance adds atmosphere (air) between the lens and the subject. Atmosphere almost always has some turbulence, turbidity or other optical disturbance. More subject distance = more optical disturbance = lower picture quality. So, for the same subject size in the frame, generally speaking longer lens = fuzzier picture.

Do NOT lower your expectations of what is achievable. Instead, raise your skills. Mainly, get a lot closer, add patience, and keep at it. Remember that for all those great photos of wildlife that you see, the photographer is only showing you his best.

 

There is an old saying that "no one sees what is on the cutting room floor."

Or as I tell my students, "no one sees your rejects." So shoot a LOT. The more you shoot the larger in absolute numbers that 5% becomes; 5% of 100 = 5, but 5% of 1000 = 50. So you have a larger pool of good pictures to pick from.