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yong54321 started following PeterStudz
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Venus with iPhone 25/06/23
yong54321 replied to PeterStudz's topic in Imaging - Smartphone / Tablets
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. -
Starsense Explorer Stand Alone Conversion
yong54321 replied to johninderby's topic in Discussions - Mounts
Here is a link to build your own starsense explorer unit. Parts from online market. -
Stacking App on iPhone - Saturn/Jupiter/Mars
yong54321 replied to PeterStudz's topic in Imaging - Smartphone / Tablets
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 -
Stacking App on iPhone - Saturn/Jupiter/Mars
yong54321 replied to PeterStudz's topic in Imaging - Smartphone / Tablets
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 -
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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Anyone know how Astronomy app (e.g. Google star map) work?
yong54321 replied to yong54321's topic in The Astro Lounge
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. -
Anyone know how Astronomy app (e.g. Google star map) work?
yong54321 replied to yong54321's topic in The Astro Lounge
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 -
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?