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Posts posted by yong54321
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Here is a link to build your own starsense explorer unit. Parts from online market.
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14 hours ago, Stu said:
Those are superb results @PeterStudz, really impressive.
I had another play with a Jupiter video after reading your post. The hardest bit seems to be getting a piece of video short enough not to crash the app, but with good content in it.
This was the best I could get…
Hi Stu
if you can send me your video, I could find out why it crashes and provide an update to videoStack. There are all kind of unpredictable reason why a video crash eg. One is a frame in the video happens to have no planet.
Please email the video to ustransit74@gmail.com
Thank you.
Best regards
yongchong
Developer
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Hi PeterStudz
Thank you very much for this post. I am very happy to see how you use these apps and produce wonderful images. Could you send the video or provide a link so that I can make it better?
yongchong
App Developer
ustransit74@gmail.com
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On 14/07/2022 at 01:09, Astro_Dad said:
Makes perfect sense - I suggested essentially same in my correspondence with Celestron. Maybe refinement will come, but for now it’s already impressive tech and works well. Not particularly familiar with SkyEye as I believe Android only, otherwise would have tried I’m sure - but principle of use is pretty clear. Thanks.
AD
It is not Android only, if you have iPhone, you can try
https://apps.apple.com/us/app/skeye-pushtocam-for-telescope/id1639616441
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On 24/11/2020 at 09:16, cdbrad said:
I'm looking for an app that will allow me to find stars/planets with my smartphone connected to my telescope. I believe SkEye is the type of app I'm looking for but its not compatible with iPhones. Any ideas on alternatives?
hi, here is a skeye alternative at iPhone
https://apps.apple.com/us/app/skeye-pushtocam-for-telescope/id1639616441
Best regards
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Hi vlaiv
Thanks very much! After reading your reply, I manage to work out the reason why N = Z - Z.U x U
I have upload my diagram in case anyone is interested.
Because |U| is also 1.
So Z.U = |Z||U|cos angle
Assuming angle is 45 degree,
Z.U = 1x1x0.707
It is 0.707 of the whole U vector.
And N = Z - 0.707U as in the diagram.
Thanks again to vlaiv.
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Hi vlaiv
Thank you very much for your answer. I have come to understand it. Just to clarify this part.
You say "If you scale celestial up with that value - len(celestial_up) cancels and you get only cos(angle) * len(z)". How does len(celestial_up) cancels when you scale celestial up with the zDotUp? It seems to me up is a vector and zDotU is a scalar...
Best regards
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Hi all,
A smart phone is able to tell you which part of the sky it is pointing to. Here is a mystery that I have pondered for quite some time, I wonder how the program in star map app work? The most interesting part is the astronomy model. Below is the crucial part of the program which is available at https://github.com/sky-map-team/stardroid
private void calculateLocalNorthAndUpInCelestialCoords(boolean forceUpdate) { long currentTime = clock.getTimeInMillisSinceEpoch(); if (!forceUpdate && Math.abs(currentTime - celestialCoordsLastUpdated) < MINIMUM_TIME_BETWEEN_CELESTIAL_COORD_UPDATES_MILLIS) { return; } celestialCoordsLastUpdated = currentTime; updateMagneticCorrection(); RaDec up = calculateRADecOfZenith(getTime(), location); upCelestial = GeocentricCoordinates.getInstance(up); Vector3 z = AXIS_OF_EARTHS_ROTATION; // (0,0,1) float zDotu = scalarProduct(upCelestial, z); trueNorthCelestial = addVectors(z, scaleVector(upCelestial, -zDotu)); trueNorthCelestial.normalize(); trueEastCelestial = Geometry.vectorProduct(trueNorthCelestial, upCelestial); // Apply magnetic correction. Rather than correct the phone's axes for // the magnetic declination, it's more efficient to rotate the // celestial axes by the same amount in the opposite direction. Matrix33 rotationMatrix = Geometry.calculateRotationMatrix( magneticDeclinationCalculator.getDeclination(), upCelestial); Vector3 magneticNorthCelestial = Geometry.matrixVectorMultiply(rotationMatrix, trueNorthCelestial); Vector3 magneticEastCelestial = vectorProduct(magneticNorthCelestial, upCelestial); axesMagneticCelestialMatrix = new Matrix33(magneticNorthCelestial, upCelestial, magneticEastCelestial); }
If you understand what is happening, please explain to me as I am in the dark for months!... Thanks in advance!
1. Why is " trueNorthCelestial = addVectors(z, scaleVector(upCelestial, -zDotu))" in the above code? So why is true North equals to upCelestial (zenith) dot product z and added to z axis? Where do this come from?
2. Why need to normalise the trueNorthCelestial.normalize(); ?
/** * Updates the astronomer's 'pointing', that is, the direction the phone is * facing in celestial coordinates and also the 'up' vector along the * screen (also in celestial coordinates). * * <p>This method requires that {@link #axesMagneticCelestialMatrix} and * {@link #axesPhoneInverseMatrix} are currently up to date. */ private void calculatePointing() { if (!autoUpdatePointing) { return; } calculateLocalNorthAndUpInCelestialCoords(false); calculateLocalNorthAndUpInPhoneCoordsFromSensors(); Matrix33 transform = matrixMultiply(axesMagneticCelestialMatrix, axesPhoneInverseMatrix); Vector3 viewInSpaceSpace = matrixVectorMultiply(transform, POINTING_DIR_IN_PHONE_COORDS); Vector3 screenUpInSpaceSpace = matrixVectorMultiply(transform, screenInPhoneCoords); pointing.updateLineOfSight(viewInSpaceSpace); pointing.updatePerpendicular(screenUpInSpaceSpace); }
3. Why is the transform in the relationship axesCelestial = T * axesPhone as described below? How is the axis of the phone related to the axis of the celestial martrix? Is this related to coordinate axis transform?
/** * The model of the astronomer. * * <p>Stores all the data about where and when he is and where he's looking and * handles translations between three frames of reference: * <ol> * <li>Celestial - a frame fixed against the background stars with * x, y, z axes pointing to (RA = 90, DEC = 0), (RA = 0, DEC = 0), DEC = 90 * <li>Phone - a frame fixed in the phone with x across the short side, y across * the long side, and z coming out of the phone screen. * <li>Local - a frame fixed in the astronomer's local position. x is due east * along the ground y is due north along the ground, and z points towards the * zenith. * </ol> * * <p>We calculate the local frame in phone coords, and in celestial coords and * calculate a transform between the two. * In the following, N, E, U correspond to the local * North, East and Up vectors (ie N, E along the ground, Up to the Zenith) * * <p>In Phone Space: axesPhone = [N, E, U] * * <p>In Celestial Space: axesSpace = [N, E, U] * * <p>We find T such that axesCelestial = T * axesPhone * * <p>Then, [viewDir, viewUp]_celestial = T * [viewDir, viewUp]_phone * * <p>where the latter vector is trivial to calculate. * * <p>Implementation note: this class isn't making defensive copies and * so is vulnerable to clients changing its internal state. * * @author John Taylor */
4. Finally the axes of the phone is given in rotation Matrix. So I wonder how is axes being represented in a rotation matrix?
I would also be happy is you can point me to where I can read up more. As I do not know where to start?
Best regards and happy star gazing!
Yongchong
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Beautiful. What power is the binocular?
Venus with iPhone 25/06/23
in Imaging - Smartphone / Tablets
Posted
Hi
I manage to download your video and stack with a newer version of videoStack for Planet app. Will inform you when this version is approved by Apple. Here you have the horns with 20 best out of 200 frames.