All you math whizzes FL and speed formula

[theantiquetiger]

I was wondering if anyone has the formula for lens movement. For example, if you are zoomed out 300mm and your handshake moves the camera 1/32 of an inch (0.07mm), how much movement occurs on the other end?

[Iguanasan]

I don't know if I'm a math whiz but I'll give it a whirl.

If you were to move it perfectly level up, down, left or right, it would only move 0.07mm regardless of how far out you are focused since everything would be perfectly parallel. The problem comes when you move it on an angle. Then I think you enter the realm of angles of elevation. Let's assume, for a moment, that you only move it straight up or down angle-wise. That is, you do not keep it parallel. You end up with this kind of movement:



So, the TAN A = 0.07 / 300 = 0.23 e-4. Clear as mud so far? If TAN A = 0.23 e-4, then A = 0.0134 (rounded off). So the angle change at the tip of the lens is 0.0134 degrees. Let's see what difference that makes. Let's say, since you have the 300mm lens out you focus down the end of an average football field which is about 100 yards and let's round it up to 100 meters to make the math a little easier since it's pretty close anyway.

100 meters = 100,000 mm. So, if we know that the focal plane is 100,000 mm away and you move the lens 0.23 e-4 degrees then the movement at the other end of the field would be x where:

TAN 0.23 e-4 = x / 100000

x = TAN 0.23 e-4 * 100000 = 0.407mm

If the focal plane is 1km away or 1,000,000 mm then x = 4mm

Let's, for fun, make the original movement of the lens 2mm. The difference is now 11mm for the football field and 110mm for the 1km distance.

If the lens moves 13mm (about half an inch) then we are looking at:

TAN (13/300) * 100,000 = 75 mm for the football field. (about 3 inches)
TAN (13/300) * 1,000,000 = 750mm for the 1 km distance. (about 30 inches )

Hmmm... I'm thinking my math sucks. I think it would be more than that but maybe not. Anyone else out there have any math skills?

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I think it's more pertinent to talk about the change in angle. If you move the camera through a 10 degree angle (10 degrees is not very much) then you would effect the football field focal point by: TAN (10) * 100000 mm or 17 meters. That's significant and what I normally would expect from what would be considered a small movement. This would have been caused by a 53mm or 2 inch movement of the tip of the lens.

[theantiquetiger]

Holy crap Iggy, you are either a genius or a BS artist, but which ever it is, that's a cool answer.

I realized my question was incorrect to begin with. 1/32" is .7mm, not .07mm (I was thinking cm).

I was just thinking about the distance of movement. Take a laser pointer and aim it 50m away. Now move the angle of the laser pointer .7mm at the back end of the pointer. That is changing the angle. How far did the red dot move 50m away?

That should be easy enough to figure out. I could probably do it. I sucked at geometry. I was an algebra/calculus type of math person.

[theantiquetiger]

Ok, just did an experiment since my last post.

I took a laser pointer and shot it about 30ft (9.1m) and made a mark.

I changed the angle by moving the laser pointer about 3/16" (4.76mm)

The laser moved 18" (0.457m)

I don't know this helps, but it might.

[Iguanasan]

Ok, just did an experiment since my last post.

I took a laser pointer and shot it about 30ft (9.1m) and made a mark.

I changed the angle by moving the laser pointer about 3/16" (4.76mm)

The laser moved 18" (0.457m)

I don't know this helps, but it might.

Genius... no, I don't think so, but thanks. I know a little trigonometry and what I can't remember, Google provides

Ok, so if you moved the laser pointer 0.457m and it was 9.1m away then change the triangle above to have the 9.1m on the bottom and 0.457m on the right hand vertical side. That means the TAN (A) = opposite side / adjacent side = 0.457 / 9.1 = 0.05. The inverse tangent of 0.05 is 2.87 or around 3 degrees. You did not move that laser pointer very much.

So, to continue the thought. A 3 degree change over the length of a football field would be

TAN (3) = X / 100

TAN (3) * 100 = X = 5 meters.

That makes sense too if you think about it. About .5 meters over 10 meters distance then it would be 5 meters over 100 meters distance. It's a simple line formula.

So, the general rule is for a given angle, A, the movement of the focal point, X, which is Y meters away would be:

X = TAN(A) * Y

A = 15 degrees, Y = 40 meters then X = TAN(15) * 40 = 10.7 meters.

You can test the math with your laser pointer if you want.

If you are 9.1 meters away move the laser pointer 10 degrees (1/9 of the way from horizontal to vertical).

A = 10, Y = 9.1, X = TAN (10) * 9.1 = 1.6 meters or about 5.2 feet.

[Marko]

I'm impressed iggy. I'm not exactly sure what the point is or the practical aspect but I'm still impressed - lol.

[theantiquetiger]

I'm impressed iggy. I'm not exactly sure what the point is or the practical aspect but I'm still impressed - lol.

I was trying to explain speed equal to/greater than FL in a beginner critique on another board. So after I explained it, I was wondering what the formula was to figure out if you moved (shook) the camera 1/32 of in inch, how much blur (movement) would it create on something zoomed in 300mm.

[Marko]

I'd be curious to see some examples if you make them...

[Iguanasan]

I was trying to explain speed equal to/greater than FL in a beginner critique on another board. So after I explained it, I was wondering what the formula was to figure out if you moved (shook) the camera 1/32 of in inch, how much blur (movement) would it create on something zoomed in 300mm.

I think it depends more on how far away have you pointed the lens. You can be zoomed in at 300mm on something that's 10 feet away to make it large in the frame or you can be zoomed in at 300mm and focused on the moon. I know when I've zoomed in on the moon you only need touch the lens a tiny bit to have the moon disappear completely out of frame.

[asnow]

I went through Engineering and as much as they teach you math they call it School of Practical Science. So the practical aspect is the rule of thumb .... speed is 1/the focal length of the lens i.e 300 mm lens is 1/300 sec. Speed can be slower with image stabilization (VR for those Nikon users)