Fun With Math

[MarkBarbieri]

OK, I'm working this all out using stuff I learned in high school several decades ago. Still, my numbers match the online calculator that I tried, so I think I've got it right. If you see any problems with my calculations, please feel free to correct me.

How Long of a Lens Do I Need?

I hear various forms of this question pretty often. It's straightforward enough to answer if you know three things - what size is the camera's sensor, what is the focal length of the lens, and how far away is the subject.

For simplicity, let's start by assuming that you are using a 35mm film camera and your subject is 100 feet away. We'll also assume that you are using a 200mm lens. The question we are trying to answer is "how wide and tall will my subject area be 100 feet away from me.

OK, 35mm isn't quite correct for the sensor size. Full frame is really about 36mm by 24mm. If you have a different size sensor, divide those dimensions by 1.3 (Canon 1D), 1.5 (Nikon/Pentax/Sony), 1.6 (Canon), 2.0 (Olympus). If you are using a point & shoot, I probably lost you by now. If not, just pretend that your using a 35mm film camera and use the 35mm equivalent focal lengths specified for your camera and not the actual focal length. Like so many other things about your camera, someone already did the work for you.

Now, the formula is simple:

Viewable_subject_width = subject_distance * sensor_width / focal_length

In our case, that's 100 feet * 36mm / 200mm = 18 feet

Viewable_subject_height = subject_distance * sensor_height / focal_length

In our case, that's 100 feet * 24mm / 200mm = 12 feet

Here's an example using a 1.6 sensor and an 18mm lens with a subject that is only 10 feet away:

10 feet * (36mm / 1.6) / 18mm = 12.5 feet wide
10 feet * (24mm / 1.6) / 18mm = 8.33 feet tall

That's it. Just multiply the subject distance by the relevant sensor size dimension and divide by the focal length.

If you want to save time and trouble, you can even memorize the ratios for your lenses on your camera. For example, using my 70-200mm lens on my 1.3x sensor, my ratios are roughly 0.4 x 0.26 to 0.14 x 0.075. So at one hundred feet, my subject size range is 40'x26' on the wide end and 14'x7.5' on the long end. So now I know that if I'm standing court side at a Rockets game, I can completely fill a landscape frame with Yao Ming when he's on the opposite end of the court. Switching to portrait mode, I can zoom out far enough to see the entire 7'6" center as long as he stays at least 18.75 feet away (7.5' / 0.4).

 

[MarkBarbieri]

I should probably add that I never do these calculations for myself. You quickly get used to how things look through your lenses. I can probably make a better guess as to what something will look like in my lenses than I can guess how far away it is. This is just something that I do when someone asks what lens they'll need at their kid's school play when they'll be 50 feet from the stage and they want a closeup of little Billy's head.

 

[jann1033]

OK, I'm working this all out using stuff I learned in high school several decades ago. Still, my numbers match the online calculator that I tried, so I think I've got it right. If you see any problems with my calculations, please feel free to correct me.

How Long of a Lens Do I Need?

I hear various forms of this question pretty often. It's straightforward enough to answer if you know three things - what size is the camera's sensor, what is the focal length of the lens, and how far away is the subject.

For simplicity, let's start by assuming that you are using a 35mm film camera and your subject is 100 feet away. We'll also assume that you are using a 200mm lens. The question we are trying to answer is "how wide and tall will my subject area be 100 feet away from me.

OK, 35mm isn't quite correct for the sensor size. Full frame is really about 36mm by 24mm. If you have a different size sensor, multiply those dimensions by 1.3 (Canon 1D), 1.5 (Nikon/Pentax/Sony), 1.6 (Canon), 2.0 (Olympus). If you are using a point & shoot, I probably lost you by now. If not, just pretend that your using a 35mm film camera and use the 35mm equivalent focal lengths specified for your camera and not the actual focal length. Like so many other things about your camera, someone already did the work for you.

Now, the formula is simple:

Viewable_subject_width = subject_distance * sensor_width / focal_length

In our case, that's 100 feet * 36mm / 200mm = 18 feet

Viewable_subject_height = subject_distance * sensor_height / focal_length

In our case, that's 100 feet * 24mm / 200mm = 12 feet

Here's an example using a 1.6 sensor and an 18mm lens with a subject that is only 10 feet away:

10 feet * (36mm / 1.6) / 18mm = 12.5 feet wide
10 feet * (24mm / 1.6) / 18mm = 8.33 feet tall

That's it. Just multiply the subject distance by the relevant sensor size dimension and divide by the focal length.

If you want to save time and trouble, you can even memorize the ratios for your lenses on your camera. For example, using my 70-200mm lens on my 1.3x sensor, my ratios are roughly 0.4 x 0.26 to 0.14 x 0.075. So at one hundred feet, my subject size range is 40'x26' on the wide end and 14'x7.5' on the long end. So now I know that if I'm standing court side at a Rockets game, I can completely fill a landscape frame with Yao Ming when he's on the opposite end of the court. Switching to portrait mode, I can zoom out far enough to see the entire 7'6" center as long as he stays at least 18.75 feet away (7.5' / 0.4).

holy crop, i should have known not to bother opening this when i saw math in the title.

 

[JR6ooo4]

holy crop, i should have known not to bother opening this when i saw math in the title.

well you see the bubbles in his avatar. He is not under water, that is hydrogen blowoff from the brain engine being run at only 67%.

Mikeeee

 

[DVC Jen]

holy crop, i should have known not to bother opening this when i saw math in the title.

I had to open it just to see if he was joking or not.

now my head hurts.

 

[tomorrowland]

I shouldn't have read that just before bedtime. That takes a fresh brain to comprehend.

 

[RainyDays]

I shouldn't have read that just before bedtime. That takes a fresh brain to comprehend.

I did read it with a fresh brain ... and cup of coffee I might add and he still lost me right after he said "For simplicity,....."

 

[MarkBarbieri]

Hmmm...sounds like there is a big demand for more of these. I'll see if I can put something together on calculating the angle of view in similar situations (that's more fun because you get to use trig functions). I could even do some stuff on calculating DOF and circles of confusion. Maybe we could even do a Photo Math contest every week.

 

[MICKEY88]

this definitely cleared something up for me..

what kind of photographer runs from a wild animal. leaving their gear behind..

 

[DVC Jen]

Hmmm...sounds like there is a big demand for more of these. I'll see if I can put something together on calculating the angle of view in similar situations (that's more fun because you get to use trig functions). I could even do some stuff on calculating DOF and circles of confusion. Maybe we could even do a Photo Math contest every week.

I think I mastered circles of confusion with this post.

 

[jann1033]

this definitely cleared something up for me..

what kind of photographer runs from a wild animal. leaving their gear behind..

the kind that doesn't want to leave their "behind" behind ( ie in the gator's mouth)?

i have to say i think i would have at least grabbed the camera, if nothing more than to use the tripod as a spear if need be
you know i actually took trig, (not by choice, had to for organic chem in college) but to me photography should be fun so that excludes having to think to much
but mark if you really wants to annoy, er i mean educate, us tackle the golden rectangle bit...i read the def. for that 3 times then figured i'd just stick to rule of thirds

 

[MarkBarbieri]

this definitely cleared something up for me..

what kind of photographer runs from a wild animal. leaving their gear behind..

Technically, I got away with my 100mm macro, which is what I had in my hand when my gator friend started his approach. The kind of photographer that runs from a wild animal leaving their gear behind is the kind that survives to post messages about their experience.

 

[handicap18]

After reading the 1st post. I got an ice cream headache.

 

[webshark3]

Disney + Prime Wide Angle Lens = Magic

But I'm sure Mark with correct my math.

 

[MICKEY88]

Technically, I got away with my 100mm macro, which is what I had in my hand when my gator friend started his approach. The kind of photographer that runs from a wild animal leaving their gear behind is the kind that survives to post messages about their experience.

but the kind of photographer that grabs his gear, particularly a sturdy tripod with camera mounted...has a weapon in hand, I'd much rather stuff my camera and tripod down the throat of a gator, than any body parts..

 

[alan]

A very useful post, Mark... But rather you than me working in those bizarre-o feet and inches...

Good luck!

regards,
/alan

 

[MarkBarbieri]

A very useful post, Mark... But rather you than me working in those bizarre-o feet and inches...

Good luck!

regards,
/alan

I hear you. I'm a big fan of the US, but our insistence on imperial units of measure is extremely annoying.

The next item(s) I'm working on for my blog (which I'll repost here) is a series on what happens when you take a picture with a DSLR. How does the image get from through the camera to your eye? How does the autofocus work? How does the metering work? How does the camera set the aperture? How does the shutter work? How does the flash work? How does the sensor work? How does the sensor data get translated to a JPG? Obviously, everything will be discussed at a very summary level as you could write a book on any of these topics. I just want to paint a picture for people of what is involved in making a picture. I think that understanding the basics will help them better understand why their camera behaves the way it does.

 

[AlienBrain]

I liked this post. It seems like I may be in the minority here when I say that Math makes things more understandable because you can quantify what you are speaking about. Thanks...

 

[mabas9395]

OK, story problem time.

I want to be able to shoot all the way down the length of a 100 yard soccer field with my 1.6 crop Canon camera and have my 5ft son fill at least 2/3 of the frame in normal landscape orientation. At the same time, another train leaves Seattle going 85mph carrying a guy with a Nikon....

The question is, what is the cheapest lens that will let me do that with a reaonable amount of IQ. I already have a 1.4x extender and want a lens that will still autofocus (app = f/5.6).

You are not required to show your work to get credit but if you can talk my wife into getting me the lens for father's day you get double points!

 

[MICKEY88]

OK, story problem time.

I want to be able to shoot all the way down the length of a 100 yard soccer field with my 1.6 crop Canon camera and have my 5ft son fill at least 2/3 of the frame in normal landscape orientation. At the same time, another train leaves Seattle going 85mph carrying a guy with a Nikon....

The question is, what is the cheapest lens that will let me do that with a reaonable amount of IQ. I already have a 1.4x extender and want a lens that will still autofocus (app = f/5.6).

You are not required to show your work to get credit but if you can talk my wife into getting me the lens for father's day you get double points!

stand at the center of the field the lens would be less expensive