focal length – How to estimate pixels per degree for an image from a camera phone using EXIF data?

This is my first time to look at metadata and I’m confused by the focal length stated in this file that a friend e-mailed to me. It was taken with a cell phone.

I’m trying to roughly estimate the angle in degrees that corresponds to a width of 400 pixels in the 4096 × 2304 pixel image. If the no zoom was used, the focal length is 3.5 mm, the diagonal of 1/3.1 inch or 8.19 mm corresponds to 4700 pixels, and using trigonometry the angle from the center to the corner is about 49.5 degrees corresponding to 2350 pixel half-diagonal, giving roughly 47.5 pixels per degree. I used (180/pi) * arctan2(8.1935/2., 3.5) = 49.5

But when I look at the metadata for the image I see Digital Zoom Ratio: 1.4.

I am thinking that the image on the phone was zoomed by hand on the touch screen, and so I need to use a different focal length for my calculation.

Should I multiply the 3.5 mm focal length by a factor of 1.4 and then use the same math, getting a new estimate of about 39.9 pixels per degree? (180/pi) * arctan2(8.1935/2., 3.5*1.4) = 39.9 Or am I missing something?

EXIF data

TIFF data General data

Estimate the rigid body transformation A, subject to T = inv (A) BA, T, and B are also rigid body transformations

Suppose we have many (possibly noisy) observations on rigid body transformations T and B, for example, drawn from SLAM path estimates.

We know that they are related by the transformation of the rigid body A, so that $ T = A -1 BA $.

How can we estimate A?

estimate – how long [roughly] Would it take to complete a Three JS project?

We are planning to create a WebGL based website. Our base would be this wonderful weekend with the Google page created by Active Theory. We have never created a project based on Three JS or WebGL before. We are aware that the page may not even have been created with ThreeJS, but I think we will stick with it. According to the client, the creative agency would also provide the 3D models.

For anyone with experience:

  1. What is the learning curve like?
  2. How long would it take to complete a project like the linked one?

Thanks in advance!

Estimate: minimize the function with log

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pr.probability – Estimate $ mathbb E_A[e^{-t|A|_2}]$, for $ t ge0 $ and random matrix m by n with iid entries with law $ N (0,1) $

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  • Please make sure answer the question. Please provide details and share your research!

But avoid

  • Ask for help, clarification or respond to other answers.
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Use MathJax to format equations. MathJax reference.

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ios: precise way to estimate the size of the output video from an AVAssetExportSession

I am building a video trimmer and exporter and I would like to display the estimated output file size for a video based on the trimmed time range and export presets. The source video I'm testing with is ~ 31MB.

the AVAssetExportPresetLowQuality and AVAssetExportPresetMediumQuality Preset estimates quite accurate. They report an estimated output size of 1.6MB and 7.8MB respectively, and the actual exported videos are basically that size.

However, all other valid export preset qualities are very inaccurate. For example, him AVAssetExportPresetHighestQuality It exports to the original file size of 31 MB, but is estimated to export 257 MB. Not even close.

Does anyone have an idea why these values ​​are so far away? Here is my sample export code, as well as the estimated and actual file sizes.

let presets = (
    AVAssetExportPresetLowQuality,
    AVAssetExportPresetMediumQuality,
    AVAssetExportPresetHighestQuality,
    AVAssetExportPreset640x480,
    AVAssetExportPreset960x540,
    AVAssetExportPreset1280x720,
    AVAssetExportPreset1920x1080
)

let formatter = ByteCountFormatter()
formatter.allowedUnits = (.useMB, .useKB)

let url = Bundle.main.url(forResource: "big_buck_bunny", withExtension: "mp4")!
let asset = AVURLAsset(url: url)

for preset in presets {
    let session = AVAssetExportSession(asset: asset, presetName: preset)!

    let output = (NSTemporaryDirectory() as NSString).appendingPathComponent(preset + ".mp4")
    session.timeRange = CMTimeRange(start: .zero, end: asset.duration)
    session.outputURL = URL(fileURLWithPath: output)
    session.outputFileType = .mp4
    session.shouldOptimizeForNetworkUse = true

    print("Estimated size for (preset): (formatter.string(fromByteCount: Int64(session.estimatedOutputFileLength)))")

    session.exportAsynchronously {
        print(session.outputURL)
    }
}

Estimated output sizes reported:

Estimated size for AVAssetExportPresetLowQuality: 1.6 MB
Estimated size for AVAssetExportPresetMediumQuality: 7.8 MB
Estimated size for AVAssetExportPresetHighestQuality: 257.6 MB
Estimated size for AVAssetExportPreset640x480: 37 MB
Estimated size for AVAssetExportPreset960x540: 54.9 MB
Estimated size for AVAssetExportPreset1280x720: 109.7 MB
Estimated size for AVAssetExportPreset1920x1080: 155.6 MB

Actual output sizes:

Actual size for AVAssetExportPresetLowQuality: 1.6 MB
Actual size for AVAssetExportPresetMediumQuality: 7.8 MB
Actual size for AVAssetExportPresetHighestQuality: 30.7 MB
Actual size for AVAssetExportPreset640x480: 27 MB
Actual size for AVAssetExportPreset960x540: 51.5 MB
Actual size for AVAssetExportPreset1280x720: 100.8 MB
Actual size for AVAssetExportPreset1920x1080: 30.7 MB

Also, it is strange that 960×540 and 1280×720 are exported in a larger size than the original, which has a resolution of 1920×1080.

video streaming: any way to estimate the bandwidth needed to use VNC?

I need to stream Linux software to Windows. Depending on your setup, I will have to stream 2.4 or 8 screens in 1080p. However, I cannot install VNC to test it myself (restrictions on my computer).
Is there a way to estimate the approximate flow I will need to stream my software?

pdes analysis: a limit Schauder estimate

According to Theorem 1.1 & # 39; in this paper we have the following estimate on classic solutions $ u in C ^ 2 ( overline {B_1 ^ +}) $ from $ – Delta u = f text {in} B_1 ^ + = B_1 cap {x _n ge 0 } $ Y $ u = 0 text {on} partial B_1 ^ + cap {x_n = 0 } $

$$ | D ^ 2u (x) – D ^ 2u (y) | le C left (r lVert u rVert_ {L ^ infty (B_ {1} ^ +)} + int_0 ^ {r} frac { omega_f (t)} {t} , dt + r int_ {r} ^ {1} frac { omega_f (t)} {t ^ 2} , dt right) tag {1} $$ $ forall , x, y in B_ {1/2} ^ {+} $ with $ r = | x-y | $ where, $ omega_f $ denotes the continuity module of $ f $ which we assume to be continuous Dini.

The proof is assumed to be similar to the proof of Theorem 1.1, which is the interior estimate. Following the lines of this test, we can obtain all the analogous estimates of the harmonic function & # 39; limit & # 39; in $ B_ {1} ^ + $ via. applying the estimates in Theorem 1.1 to the odd extensions of the harmonic functions to the full ball $ B_1 $. But in the last step (following the lines of the equation $ (1.13) $ in the article) it seems we need the estimate $$ | D ^ 2u_0 (x) – D ^ 2u_0 (y) | le C lVert u rVert_ {L ^ { infty} (B_1 ^ +)} | xy |, , forall , x, y in B_ {1/2} ^ + tag {2} $ PS where, $ u_0 $ satisfies $ – Delta u_0 = f (0) $ in $ B_ {1} ^ + $ Y $ u_0 = 0 $ in $ partial B_1 ^ + cap {x_n = 0 } $ where, $ C $ is supposed to be independent of $ f $.

I can't seem to get the estimate $ (2) $ for half a ball.

In case of internal estimation we can consider $ v_0: = u_0 – frac {f (0)} {2n} (1 – | x | ^ 2) $ which is harmonic in $ B_1 $ and writes begin {align *} | D ^ 2u_0 (x) – D ^ 2u_0 (y) | = | D ^ 2v_0 (x) – D ^ 2v_0 (y) | & le r lVert D ^ 3v_0 rVert_ {L ^ { infty} (B_ {1/2})} \ & le r lVert v_0 rVert_ {L ^ { infty} ( partial B_ { 1})} = r lVert u_0 rVert_ {L ^ { infty} ( partial B_ {1})} end {align *} which shows the inner analogue of $ (2) $ using only the gradient estimate for the harmonic function $ v_0 $.

But a similar approach doesn't seem to be working for the borderline case (for example, considering $ u_0 – frac {f (0)} {2} x_n ^ 2 $ which is harmonic and applies an odd extension to this.)

Any help is appreciated. Thank you.

postgresql – Is there any possible way to get a progress estimate to ALTER TYPE OF COLUMN in Postgres?

I know there is currently no officially supported way, but I have almost two 100% CPU hours on a 15,000,000 row table changing an INTEGER to SMALLINT and I have no idea how long this will continue. Is there a file I can see grow on the file system? Do you rewrite the table instead or create a new file? Is there a way to pause this so that I can run a benchmark on the same machine with fewer rows, so I can get a row per second estimate for this operation?

Estimate of the sum $ sum_ {1 leq x, y leq n} x mathrm {lcm} (x, y) ^ {- 1} $

I'm looking for approximations, or a closed form, if available, for the sum

$ S (n, a, b) = sum_ {1 leq x, y, leq n} frac {x ^ a} { mathrm {mcm} (x, y) ^ {b}} $

where $ mathrm {mcm} (x, y) $ is the least common multiple of integers $ x, t $ Y $ a, b $ They are positive quantities. I am particularly interested in $ a = b = 1. $ For this case, the numerical evidence suggests
$$
S (n, 1,1) = O (n log n)
$$

I could celebrate. In particular, I wonder if by using the technique in the answer to this question here, one could get (like $ n rightarrow infty $), Leaving $ a, b downarrow 1, $ an estimate in terms of zeta functions. In that question the upper limit
$$
S (n, 0, b) leq frac { zeta (b) ^ 3} { zeta (2b)}, quad b> 1
$$

is derived leaving $ n rightarrow infty. $
Any pointer, comments welcome.