magnetometer

Compute rotation matrix using the magnetic field

╄→гoц情女王★ 提交于 2019-12-03 11:21:35
问题 In get rotation matrix value it contains public static boolean getRotationMatrix (float[] R, float[] I, float[] gravity, float[] geomagnetic) Here how can i calculate the float[] gravity ? I found a sample of code where it calculate the orientation using both Accelerometer and Magnetic field boolean success = SensorManager.getRotationMatrix( matrixR, matrixI, valuesAccelerometer, valuesMagneticField); if(success){ SensorManager.getOrientation(matrixR, matrixValues); double azimuth = Math

Android: Algorithms for SensorManager.getRotationMatrix and SensorManager.getOrientation()

回眸只為那壹抹淺笑 提交于 2019-12-03 09:05:13
To get orientation as a from of euler angles (e.g., pitch, roll, azimuth) in Android, it is required to execute followings: SensorManager.getRotationMatrix(float[] R, float[] I, float[] gravity, float[] geomagnetic); SensorManager.getOrientation(float[] R, float[] orientation); In the first one, I realize that it uses a kind of TRIAD algorithms; Rotation matrix (R[]) is composed of gravity, geomagnetic X gravity, gravity X (geomagnetic X gravity) --- X is cross product. See codes below: float Ax = gravity[0]; float Ay = gravity[1]; float Az = gravity[2]; final float Ex = geomagnetic[0]; final

Getting magnetic north from a xyz magnetometer (iPhone 3.0)

萝らか妹 提交于 2019-12-03 05:15:36
问题 I'm actually reposting a question from the iphone development boards. I didn't ask it, but I found it when trying to find its answer. Nobody had yet responded there, so I thought I'd try posting it here. Thanks in advance for any help! Does anybody know any specifics about the algorithm of computing the magnetic heading from the raw xyz values of CLHeading? I am sure there must be a general approach to this that works for different magnetometers as well. 回答1: If you don't want to use the

Compute rotation matrix using the magnetic field

浪子不回头ぞ 提交于 2019-12-03 00:50:46
In get rotation matrix value it contains public static boolean getRotationMatrix (float[] R, float[] I, float[] gravity, float[] geomagnetic) Here how can i calculate the float[] gravity ? I found a sample of code where it calculate the orientation using both Accelerometer and Magnetic field boolean success = SensorManager.getRotationMatrix( matrixR, matrixI, valuesAccelerometer, valuesMagneticField); if(success){ SensorManager.getOrientation(matrixR, matrixValues); double azimuth = Math.toDegrees(matrixValues[0]); double pitch = Math.toDegrees(matrixValues[1]); double roll = Math.toDegrees

Getting magnetic north from a xyz magnetometer (iPhone 3.0)

徘徊边缘 提交于 2019-12-02 18:33:12
I'm actually reposting a question from the iphone development boards. I didn't ask it, but I found it when trying to find its answer. Nobody had yet responded there, so I thought I'd try posting it here. Thanks in advance for any help! Does anybody know any specifics about the algorithm of computing the magnetic heading from the raw xyz values of CLHeading? I am sure there must be a general approach to this that works for different magnetometers as well. cube If you don't want to use the trueHeading value, and assuming that the compass lies perfectly horizontal, the heading can be computed

Getting magnetic field values in global coordinates

半腔热情 提交于 2019-12-01 13:14:29
For an Android application, I need to get magnetic field measurements across the axis of global (world's) coordinate system. Here is how I'm going (guessing) to implement this. Please, correct me if necessary. Also, please, note that the question is about algorithmic part of the task, and not about Android APIs for sensors - I have an experience with the latter. First step is to obtain TYPE_MAGNETIC_FIELD sensor data ( M ) and TYPE_ACCELEROMETER sensor data ( G ). The second is supposed to be used according to Android's documentation, but I'm not sure if it shouldn't be TYPE_GRAVITY instead

CMDeviceMotion returns 0 values for magnetic field

大城市里の小女人 提交于 2019-12-01 05:43:31
I am developing iOS app with compass functionality. I have tried to use CMMagnetometerData updates which give uncalibrated, but normal results. After that I tried to get CMDeviceMotion updates which turned out to give always zero magnetic field with CMMagneticFieldCalibrationAccuracyUncalibrated accuracy. The only device I have is an iPad, so can't test on others. May be field is zero because sensor is not calibrated, but I could not find any way to perform calibration. How to fix that? UPDATE: Here is suggested to use startDeviceMotionUpdatesUsingReferenceFrame:toQueue:withHandler: , however

Which iOS class/code returns the magnetic North?

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-30 09:14:52
I want to get the device's deviation from the magnetic North in degrees, and use that value in some code I'm writing. I don't want to use the device's location services and therefore I'm not interested in getting the True north but rather the Magnetic North (using only the device's magnetometer). Which class (or coding process.. ) could provide me with that value (solely relying on the magnetometer) ? the CLLocationManager class and its properties rely on Location Services being enabled/available where as the Core Motion framework with its CMMagnetometerData class provides us with the

finding orientation using getRotationMatrix() and getOrientation()

岁酱吖の 提交于 2019-11-29 14:37:32
问题 Im trying to get the direction of the vector pointing out of the camera, with respect to magnetic north. I'm under the impression that I need to use the values returned from getOrientation(), but I'm not sure what they represent. The values I get from getOrientation() don't change predictably when I change the orientation of the phone (rotating 90 degrees does not change values by 90 degrees). I need to know what the values returned by getOrientation() mean. What I have so far is written

Which iOS class/code returns the magnetic North?

心已入冬 提交于 2019-11-29 14:04:24
问题 I want to get the device's deviation from the magnetic North in degrees, and use that value in some code I'm writing. I don't want to use the device's location services and therefore I'm not interested in getting the True north but rather the Magnetic North (using only the device's magnetometer). Which class (or coding process.. ) could provide me with that value (solely relying on the magnetometer) ? the CLLocationManager class and its properties rely on Location Services being enabled