Size lens to get for medium to long range recording.

You will need a long lens. For an example, the field of view of a 400mm lens, on a full frame camera is about 32 feet. A five foot long deer would be 1/6 the width of the frame. Your camera has a 2X crop factor so the field of view for a 400m lens would be about 16 feet at 100 yards. The deer would be about 1/3 of the width of the photo.

It is only in the movies that a person can take a shot of a person 100 yards away and get a head and shoulders photo.

To make a photo, similar in framing to the one below and without any cropping in Post Production, you will need a 500mm lens if the deer are at 100yards: if the deer are at 200yards then you will need a 1000mm lens.

So, I think you should look at the longest zoom you can afford.

WW

The astronomical crowd considers 50mm to be magnification 1. This is arguable!. If you accept this (I do), then mounting a100mm delivers 2X and a 500mm delivers 500 ÷ 50 = 10X.

I have pondered this rule of thumb and I discovered it is based on a 50mm lens mounted on a 35mm film camera. The dimensions of the film frame are 24mm by 36mm with a diagonal of 43.3mm. The photographic community rounds this value up to 50mm.

The basis of this rule-of-thumb is: Magnification 1 occurs when you mount a lens with a focal length that is approximately the same as the diagonal measure.

Now you should know that the photographic crowd considers a 50mm mounted on a 35mm film camera delivers a “normal” perspective -- thus the 50mm is a “normal” lens. Actually, mounting a lens with a focal length equal to the diagonal measure of the film frame or digital sensor, delivers “normal”. Such a lash-up delivers a 45° angle of view, camera held horizontal (landscape).

Find the specifications for your camera’s imaging chip and calculate the diagonal measure of this rectangle. Mount a lens with this focal length and call it magnification 1.

As an example: Your camera is a Lumix G7. . This format measures 13mm height by 17.3 mm length. The diagonal measure of this rectangle is approximately 21.64mm. If you mount a 300m lens, the magnification delivered is 300 ÷ 21.64 = 14X.

OK for this math: Now the rest of the story – Your camera sports a miniature size sensor; thus the resulting image is practically worthless unless magnified. You view this image on a computer screen or you make a paper print. Both will be enlargements. You must take into account the degree of magnification applied to the final image to be viewed. Find a target of known length. An automobile will do; just measure its length. Now step away and shoot pictures. You can use your zoom or fixed focal lengths. As you shoot, record the camera’s focal length and your distance to the subject. Procure a clear plastic ruler. Display the images of the cars on your computer screen. Divide the length of the car’s image on the screen into the known length of the car. That math delivers the total magnification applied by the camera and the viewing method.

"Display the images of the cars on your computer screen."

I assume that means reduce the size of the image to fit on the screen first. My laptop is 1366 x 768 pixels. The camera has a max resolution of 4592 x 3448 pixels. If I were to put that on my screen only about one third of the picture would show at a time.

There's no getting away from maths if you want to find out exactly how big a subject will appear with a certain lens.

It's all about angles.

Given the 17.3mm (horizontal) width of the G7 sensor, a 300mm lens gets you an angle of view of 3.3 degrees. The tangent of this angle, multiplied by the subject distance, will give you the subject width covered.

Luckily this complicated maths is dead easy with the aid of a calculator.

So 100yds x Tan3.3 = 5.77yds = 17.3ft.

In other words you'd need a subject over 17ft wide to fill the horizontal frame at 100 yards. A 6ft deer would only be just over 1/3rd of the frame wide.

With a 100mm lens it would only be about 1/9th of the frame wide - pretty vanishingly small.

As Bebu says, in order to get good wildlife pictures, improving your woodsmanship and stalking skills will get you better pictures than a longer lens.

Joe, check your angles. The correct result would be obtained by.

2 x (100 yards x tan 1.67 X 3) = 17.49 feet. Just to be very accurate.

There's no getting away from maths if you want to find out exactly how big a subject will appear with a certain lens.

It's all about angles.

 

Given the 17.3mm (horizontal) width of the G7 sensor, a 300mm lens gets you an angle of view of 3.3 degrees. The tangent of this angle, multiplied by the subject distance, will give you the subject width covered.

 

Luckily this complicated maths is dead easy with the aid of a calculator.

 

So 100yds x Tan3.3 = 5.77yds = 17.3ft.

 

In other words you'd need a subject over 17ft wide to fill the horizontal frame at 100 yards. A 6ft deer would only be just over 1/3rd of the frame wide.

 

With a 100mm lens it would only be about 1/9th of the frame wide - pretty vanishingly small.

 

As Bebu says, in order to get good wildlife pictures, improving your woodsmanship and stalking skills will get you better pictures than a longer lens.

I've seen many bird pictures taken with very long lens and the other day I saw on the web some pictures made by someone who fly along with the birds witha a wide angle lens. The images give quite a refreshing perspective.

Joe, check your angles. The correct result would be obtained by....

I did exactly as you suggest.

However, at such small angles the difference between 2x the half-angle, and the full angle tangent, is small enough to ignore.

The difference being an angle of:

3.303141454 degrees for 2x the half-angle

And 3.300401423 degrees as the arctan of 17.3/300.

You're quibbling over 0.0027 degrees?

When the sensor width and lens focal-length dimensions are purely nominal figures anyway?

Joe. You are quite right, that is quibbling on my part. At 3 degrees the hypotenuse and adjacent leg are practically the same length at 90 degrees and 87 degrees at the opposite leg. On larger angles, such at 45 degrees, there is a marked difference. At 100 feet 45 degrees would show 100 feet field of view. The correct method 2 x (tan 22.5 x 100) would give the correct field of view of 82.8 feet. A significant difference. I just wished to point out the correct method. I often did it incorrectly in the past.

For a very significant difference look at an 18mm lens with a 75 degree angle of view at 100 feet.

Incorrect = tan75 x 100 = 373 feet

Correct 2 x (tan 37.5 x 100) = 153 feet

Just a observation. If you watch the Nature and sometimes NOVA shows on PBS where they show professional wildlife photographers in action, you will see that these people are not messing around much with any puny lenses under 500mm, and usually their lenses are much longer than that.

My wildlife photography these days (such as it is) is often done with a 100-400mm lens on a APS-C body (equivalent to a 160-640mm in my case). Even with that, cropping is necessary for those "head and shoulder" shots.

Canon 50D with 100-400mm at 400mm setting (~640mm) - uncropped

And you thought your dog shed a lot in the early Spring!

If you're hunting in the woods a long telephoto won't help you. How far can you see? 100 yards? The other problem is the weight. Do you want to carry around a heavy camera bag? Third, what do you intend to do with these pictures? Are you blowing them up to 30" or just putting them on-line?

You may find a P&S like a Panasonic's ZS200/TZ200 is a 20MP 1"-sensor compact with range of let's say 25mm-360mm is good enough. It will fit in your shirt pocket and weigh practically nothing. Sony makes the best the RX100VI but it's about three times the cost and the range is up to 200mm. (I have a SOny RX100iv but it's zoom is only 24-70mm) 1" sensors are great for small blowups and for showing on my 75" 4K TV.

Here's a review comparison of similar small cameras with nice zooms.

Panasonic Lumix DC-ZS200/TZ200 Review

Nothing like getting up on a cold and frosty morning and heading out to track down the wild and elusive game. Oh, the thrill of it all.

http://jdainis.com/wild.jpg

As a hunter and a photographer my thought is, make up your mind what you want to be serious about. You can't do justice to both. If you are out there with the proper photo equipment it's going to get in your way if hunting is your prime goal. My advice: If you are out to bag a deer for your freezer, take along something light weight and compact for your trophy photo. If you want great deer photos doing what deer do, take something at least 500mm and leave you shooting iron behind. (do you really want blood all over that nice lens?)

The only good hunting photos I have ever seen were taken by a photographer who was with the hunter as a photographer only. Do you have a buddy who could fill that position?

Joe. You are quite right, that is quibbling on my part. At 3 degrees the hypotenuse and adjacent leg are practically the same length at 90 degrees and 87 degrees at the opposite leg. On larger angles, such at 45 degrees, there is a marked difference. At 100 feet 45 degrees would show 100 feet field of view. The correct method 2 x (tan 22.5 x 100) would give the correct field of view of 82.8 feet. A significant difference. I just wished to point out the correct method. I often did it incorrectly in the past.

 

For a very significant difference look at an 18mm lens with a 75 degree angle of view at 100 feet.

 

Incorrect = tan75 x 100 = 373 feet

 

Correct 2 x (tan 37.5 x 100) = 153 feet

There's actually no need to even involve the tangent function if you accurately know the subject distance and the sensor dimensions.

By similar triangles; the subject area is the same as the sensor area multiplied by 1/m (image magnification), and for 'normal' subject distances, magnification = focal-length divided by subject distance.

Therefore subject area = sensor-size x subject distance/focal length.

In the case above 17.3mm x 100yds/300mm.

300ft = (300x12x25.4)mm = 91,440mm

91,440/300 = 304.8

304.8x17.3 = 5273.04mm

And 5273.04mm = 17.3 ft..... exactly

This is obvious if you cancel out the 300ft (100yds) with 300mm to start with. All you then need to do is substitute feet for millimetres in the sensor width dimension.

Anyway, that still doesn't take into account the actual, as opposed to nominal, focal length. Nor any variation of focal length with focussing distance, nor any geometrical distortion in the lens.

The maths only gets you close. It's not an exact science unless all the lens parameters are known, and that would just be a mind-boggling and pointless exercise.

Mostly, close enough is good enough.;)

Joe. You are quite right, that is quibbling on my part. At 3 degrees the hypotenuse and adjacent leg are practically the same length at 90 degrees and 87 degrees at the opposite leg. On larger angles, such at 45 degrees, there is a marked difference. At 100 feet 45 degrees would show 100 feet field of view. The correct method 2 x (tan 22.5 x 100) would give the correct field of view of 82.8 feet. A significant difference. I just wished to point out the correct method. I often did it incorrectly in the past.

 

For a very significant difference look at an 18mm lens with a 75 degree angle of view at 100 feet.

 

Incorrect = tan75 x 100 = 373 feet

 

Correct 2 x (tan 37.5 x 100) = 153 feet

You're right as the formula is applicable only to a right triangle so you have to split the angle in half as 2 right triangles.

It's always best to do something the right way, which works every time.

Yes, focal lengths as written on lenses etc. are nominal. That's no reason to introduce yet another source of error by also using incorrect maths.

That's no reason to introduce yet another source of error by also using incorrect maths.

And yet my rough 'n' ready calculation came up with 17.3ft (correct), while for some reason James' method arrived at 17.49ft (inaccurate).

You're right as the formula is applicable only to a right triangle so you have to split the angle in half as 2 right triangles.

What formula?

The half-angle method is only applicable if you've already been given the angle-of-view of the lens*.

If you only know the sensor size and focal length, then you have to work out the tangent for yourself.

Say we have a 20mm lens on a 24mm wide sensor. Dividing 24 by 20 to get a tangent gives exactly the same result as dividing 12 (the half sensor-width) by 20 and then multiplying by 2.

So why walk all around the block to get back where you started?

* The published angles-of-view for lenses are usually the useless diagonal angle. And how often do you tip the camera diagonally to get all of a subject in frame?

For the sake of completeness, here's a diagram showing why working with half-angles is a waste of time, if all you want to know is the subject width/height covered.

X is the unknown subject dimension

D is the subject distance

S is the sensor width or height

F is the lens effective focal length

You can see by the theory of similar triangles that X will be the same in both cases, but it's much easier to cut to the chase and use the simpler formula on the right.

Incidentally, the right hand diagram is the same as for a shifted lens.

NB. The angles involved aren't the same in both cases, but that's irrelevant if all we need to know is the subject height or width.