yong54321

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.

Here is a link to build your own starsense explorer unit. Parts from online market.

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

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

It is not Android only, if you have iPhone, you can try https://apps.apple.com/us/app/skeye-pushtocam-for-telescope/id1639616441

hi, here is a skeye alternative at iPhone https://apps.apple.com/us/app/skeye-pushtocam-for-telescope/id1639616441 Best regards

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.

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

Beautiful. What power is the binocular?