How to calculate correct shutter speed

<p>Good evening<br>

Sorry if I'm posting in wrong forum but I have a little problem which I thought would be easy - apparently not! And I'm totally dyslexic when it comes to maths...<br>

A train is travelling at 70mph towards the camera (classic 3/4 side view). The distance between the front of the train and the camera lens at the time of capture is about 12 feet.<br>

How does one work out the "correct" shutter speed mathematically in order to freeze motion?<br>

Is there an easy way or does it involve something like calculus?<br>

I ask this question because I am fed up of relying on guesswork. Now I suspect various factors come into play and somebody is going to tell me that it depends on the aperture and the lighting etc...etc...and why don't I just set the camera to AV mode?. Well, the camera doesn't know if the subject is moving or not so it gets it wrong 99% of the time in AV mode - this is why I use manual mode. But I often get it wrong too in the effort to use as slow as speed as possible without wasting light or getting shallow DOF.<br>

A good quality explanation/formula would be most helpful please<br>

robert7110a<br>

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<p>A few observations, in no particular order:<br /><br />1) The camera can't get it wrong an Av mode. It's giving you exactly the shutter speed it has to, in order to make a decent exposure based on the other two things you've told it can't change: aperture and ISO setting. Speaking of ISO...<br /><br />2) You don't mention ISO in any of your remarks. Are you aware you can change it? Consider how you intend to actually use the image ... web display? VERY large prints? Different output formats can help to dictate what sort of high ISO degradation you can tolerate, and this how sensitive you can make the camera, and thus how shutter speed and aperture are going to interplay in a given amount of light.<br /><br />3) You're not mentioning anything about the time of day you're shooting. <br /><br />4) You're not mentioning how you're compsing the shot, and how much of the rest of the train is in the frame and needs to be in focus.<br /><br />5) At 12 feet, you're well within range of freezing the front end of that train by using flash. <br /><br />6) Are you shooting digitally? Have you considered just cranking up ISO (for experimental purposes), and then shooting in Tv so that you can directly answer your own question, empirically? Because just how frozen is frozen is actually rather subjective. You can test a range of shutter speeds on the entire train as it passes by. Just keep shooting at the same spot, and you can judge freezy-ness by looking at elements on each car - you don't need the front of the train to tell you what's going on. But with the whole train going the same speed, you have a perfect laboratory for varying the shutter speed from 1/60th through 1/2000th to see what you like given the focal length and output format you'll be using.</p>

<p>Matt<br>

The camera "can" get it wrong on AV mode. How does the camera know whether my subject is moving or not? Exactly.<br>

So I set Tv mode and a shutter speed of 1/500, The camera gets the aperture wrong. But this wasn't my question<br>

Otherwise, assume the following are constants...focal length 100mm, ISO 400 and bright sunny day,<br>

I want the whole train to be in acceptable focus with no motion blur (and yes I do understand that I will lose sharpness towards the rear of the train which is why I mention acceptable focus) <br>

How do I calculate the correct shutter speed mathematically assuming I know the train speed and distance between the train and the camera?</p>

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<p>Well at 70mph, the train is moving at 102 feet per second. At 1/250 sec shutter x sync or travel speed, the train is going to move approx 4.8 inches during the exposure. If you aren't panning, you will have a lot of smear from the top to the bottom of the frame if you are only 12 feet from the train. Now your 3/4 view will cut that down some, but you need to be a lot further away to reduce the angular velocity of the train so you can pan fast enough so the smear doesn't show in the image.</p>

<p>One factor is important. You can never perfectly freeze a moving subject. There will always be a movement but you will need to determine how much of the movement is acceptable as freeze. <br>

You said the distance is 12 ft. but the lens focal length is an factor here.</p>

 

<p>Ok, a 100mm lens and 12 feet from the train, and the whole train in the image? Something just doesn't add up when you mention AV mode. With that setup, 105mm lens on a full frame 35mm camera you have a field of view approx 2x3 feet at 12 feet.</p>

<p>I'm using manual mode (not AV or TV) and I'm not panning....</p>

<p>Robert: Av can't be wrong, and Tv can't be wrong, but <em>choosing to use</em> Av or Tv can be wrong. In either case, the camera is simply doing what you tell it to. In either mode (Av or Tv), <em>you</em> are determining both shutter speed and aperture by shooting in a given amount of light and using the ISO that you select. The shutter speed or aperture that the camera derives isn't wrong, it just is what it is because of the conditions that <em>you</em> create. The camera is no more wrong in either case than the laws of physics are wrong. <br /><br />Since you already know you want a certain amount of DoF (just use a DoF calculator <a href="http://www.dofmaster.com/dofjs.html"><strong>like this one</strong></a>), and you can readily determine the appropriate shutter speed by looking at results as you shoot with a given focal length, working distance, <em>and print/display method/size</em>, there are really only two things you can change to get the results you want:<br /><br />1) ISO setting. Make the camera sensitive enough to allow you to use the shutter speed and aperture you know you want.<br /><br />2) The light. Either add light (use this site's search tool and search on the phrase "Kent Staubus trains" - you'll see many threads involving trains and light... that's Kent's <em>thing</em>), or wait for a time of day when the available light is sufficient for your stopped-down lens, high shutter speed, and not-too-high ISO setting.</p>

<p>Thanks Matt but I think you're deviating away from my original question. Yes, one can calculate DOF using a DOF calculator but this wasn't my question. How can a camera calculate the "correct" shutter speed that "I need" rather than basing it on ISO and aperture? I will argue that AV "does" get it wrong.</p>

<p>It can't.</p>

 

<p>In this situation, you have to figure it out by trial and error, or mathematics. Plus the info you have given just doesn't fit the scenario of a photo of a train, unless all you want is a picture of the coupler on the front of the engine.<br>

What camera are you using?</p>

 

<p>If you're standing 12 ft from a train moving at 70mph you're not thinking personal safety or the well being of your gear. The turbulence created by a train moving that fast at close distances can propel small dust particles into your lens which have the propensity to do some real damage. Since you mentioned using a 100mm lens here as well....something just doesn't add up. Maybe you meant a 10mm lens, or standing 120 feet from the train.</p>

<p>Robert: The DoF calculator is to establish what aperture you need in order to get what you want in focus, given the distance between you and the subject, the format of camera you're using, and the focal length of the lens you're using. That establishes <em>one</em> aspect of the exposure. One thing that constrains you, because of <em>your</em> desire to get X amount of the train in focus. So, using the DoF calculator, or doing it by experience/seat-of-the-pants is one step in getting where you want to be.<br /><br />The next step is settling on a shutter speed. As mentioned above, there are more variables involved than you're talking about. Can you address the issue of how you'll be using the images? Again: large prints? Display online? Half a magazine page? Right there, you could be looking at the difference between 1/250th and 1/1000th - without even getting into working distance, angular velocity, and focal length, and the rest. If you're shooting digital, you've got exactly what you need, with one pass of one train, to run a test (shooting in Tv, and working your way quickly through half a dozen shutter speeds, letting the camera adjust aperture for you just for that test, to keep the exposure tolerable). <br /><br />That's the fastest way to establish what shutter speed will work for:<br /><br />1) Your working distance.<br />2) Your focal length.<br />3) Your intended output format/display size.<br />4) Your ISO tolerance (a function of the camera you're using, and - again - output format)<br />5) Your choice of aperture, established above.<br /><br />And that's it on your requirement. You may find that you <em>can't service</em> your requirement with your existing gear, based mostly on high ISO performance, and you just have to wait for strong daylight.<br /><br />As for the pedantic "get it wrong" issue. No. Automated priority modes (shutter <em>or</em> aperture) are based on the light, not on your intentions and what you know about the subject matter and its movement. Those modes can't get anything wrong, because you're only using them when you know exactly what they're going to do. Your camera will be telling you, as you compose, precisely which shutter speed or aperture it finds necessary to match up with the complementary setting you've chosen (in the context of the ISO setting you've already set). The camera isn't wrong, if the shutter speed is too slow or the aperture is too wide ... that's entirely on the operator of the camera.<br /><br />If you use the methods discussed above, you'll already know what your boundaries are when using one of those automated methods, and won't get into trouble by passing your threshold of pain in either area.</p>

<p>Here's the stop-action shutter speed formula found in the <em>Kodak Professional Photoguide</em>, 5th ed., 1995 (written for film cameras, but the principle is the same for digital):<br>

Exposure time in seconds (shutter speed) =<br>

(distance in feet x direction factor) / (875 x speed in mph x lens magnification)<br>

where:<br>

lens magnification = focal length of lens used / normal focal length for the camera<br>

direction factor: motion across field = 1<br>

motion at 45° =2<br>

motion toward or away from camera = 4<br>

According to the examples given, if the train is 12 feet from your camera and is traveling at only 10mph, the shutter speed at 45° would be 1/500th sec.; at 25 mph, the shutter speed would be 1/1000th sec. If the train is going faster (50 mph), to stop the motion, you will have to back up to at least 25 feet and shoot at the same shutter speed (1/1000th).<br>

Depending on the available light, you will probably have to boost your ISO pretty high to get an aperture with sufficient depth of field.</p>

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<p>Calculators Gone Wild<br>

Exercise in futility –</p>

<p>Given your stated conditions and your desire to “freeze motion” ...<br>

Let’s make some gross assumptions. </p>

<p>Coming from the engineering world of office copiers, an ancient rule of thumb is that the human eye at reading distances cannot resolve better than a 0.002” toner dot on white paper at “normal reading distance”. Wikipedia and other locations indicate that the limit of human acuity at near distances is 30 arc seconds, which translates to 0.002” at a viewing distance of 16 inches. So, just maybe old rules of thumb and Wiki seemingly agree. Let’s presume that here and figure you are printing in landscape form on an 11” wide sheet.</p>

<p>Therefore, figure then if at a 16” viewing distance you can resolve 500 dots/inch for 11” paper, then you need to have 5500 “visible dots” as mentioned above. Presume you use a full frame Nikon D3X (6048 horizontal pixels). Crudely speaking, that means have only 9% more pixels than you need, so you can’t afford to smear any pixels. (So we are going for "near zero" blur/smear/object movement.)</p>

<p>At 12 feet from a train surface with the D3X and a 100mm lens, your field of view is 4.32 feet. Let’s presume you are at a 45 degree angle to the train, the effective speed then is 49.8 mi/hr (73ft/sec) across your field of view. Using proportions, 4.32ft/5500 = 0.0008ft. At 49.8 mi/hr the train moves that 0.0008ft in 1/100000sec. In order to have zero blur, your shutter speed would be that! Any slower is a compromise. I have no rational way to calculate what is “acceptable” versus what is “perfect, frozen motion”.</p>

<p>So I guess what I am implying is that you want your shutter speed to be maxed out as much as possible … DOH ….</p>

<p>My apologies to all for the presumptuousness of this example.<br>

Jim</p>

<p>I think the Kodak's formula the Bob posted is a simple and good one. And you would need around 1/5000 or so shutter speed. </p>

<p>I think you need to tie a damsel to the rails to force the train to stop so you can take your picture.</p>

<p>Um I tried to post response but it didn't work for some reason. Thanks especially to Bob Sunley for suggesting the mathematics (this is what I need...so example please) and Ben Evans for the link to the formula.<br>

I use (mainly) a 5D and 70-200mm lens (at around 100mm for trains). Today the light was good so I experimented by upping the ISO to 640. With a polariser filter attached, I still managed to freeze a 70mph train at 1/850th sec. Be interesting to see what a 15 x 10 print will look like...</p>

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<p>I think you need to tie a damsel to the rails to force the train to stop so you can take your picture.</p>

</blockquote>

<p>The op could attach a 12 ft boom with a chair to the front of the train. With that the OP can use just about any shutter speed and having a great panning effect.</p>

<p>I am with Jim – perhaps mine is even a more simplistic approach.<br /> You need to use the fastest shutter speed you have: and that will NOT be fast enough, I think.</p>

<p>If you want a simple visual here is one below (speed vector diagram - right angle triangle)<br /> A ¾ Profile is about an angle of 60° - (Jim used 45°)<br /> At 60° the <em><strong>relative speed across the camera</strong> </em>is a bit slower than Jim’s example.<br /> You don’t need a lot of calculus, there are 63million trigonometry calculators on the www –</p>

<p><img src="http://d6d2h4gfvy8t8.cloudfront.net/15224713-md.jpg" alt="" width="680" height="486" /></p>

<p>Now we know that (for Jim’s example) the train is speeding ACROSS the horizontal of the camera at a relative speed of about 50mph, and in my example it has a relative speed ACROSS the camera of about 35mph.</p>

<p>So, to answer how to calculate this really simply - with that information we have witha simple trig calculator - just jump out on the footpath (“pavement” if you are in the USA) and get a friend to drive down the street at 50mph and then also 35 mph and set the camera at 12ft from the car which is passing.</p>

<p>It will be apparent that 1/8000s is still pretty slow for either speeding car <strong><em>at that shooting distance,</em></strong> no matter what lens you choose to use.<br /> (Assuming no Panning of the camera)</p>

<p>However if you mean by “Classic ¾ Side View” the train coming at 30° to the camera - the the relative speed across the camera is even fast than 50mph – more like 60mph.</p>

<p>WW</p>

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<p>Today the light was good so I experimented by upping the ISO to 640. With a polariser filter attached, <strong><em>I still managed to freeze a 70mph train at 1/850th sec.</em></strong> Be interesting to see what a 15 x 10 print will look like...</p>

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<p>OK.<br>

I posted my reply - but I did not notice that you have posted this.<br>

So you froze the movement of a train which was <strong>travelling at 70mph</strong> and at <strong>an angle to the camera</strong> with the camera <strong>at shooting distance of 12ft</strong> . . . from the train – and you did NOT pan the shot?<br>

I do not understand how that is possible.<br>

Or am I missing something?</p>

<p>WW</p>

 

At 60 MPH the train is moving at about 90 feet per second. At 1/850 second shutter speed, from the time the shutter opened to the time the shutter closed, the train would have moved about 1-1/4 inches. Sure seems like that would have been some blur.

<p>Seeing how the 5D has an X sync speed of 1/200 sec, there is going to be a substantial amount of vertical skew in the image as well if you aren't panning the shot.</p>

<p>↑<br>

In keeping it simple, I didn’t even begin to get into that area of distortion . . . creating the “leaning train”.<br>

But yes – another aspect of it all.<br>

I just noticed that you brought that up earlier in your first post – an important point.<br>

I am confused by the OP’s claims, which appear to me to be technically impossible results.</p>

<p>WW</p>

<p>#nevermind</p>

<p> </p>