I need help regarding two rotations at the same time, on the same axis. One vector is approaching 90 degrees. When I press up, it should smoothly approach the angle 45 degrees above it, such that HOLDING up will still be going down. Here is a rough sketch of what I mean: (black = vector without up, red = vector approaching 45 above black)
Please refer the official documentation on
Quaternion.Lerp here
Quaternion.Slerp here
Choose whichever suits your needs.
Related
I did some research on OpenPose and the output is x and y coordinates with confidence point. x and y coordinates are good for detecting up, down, left, and right movements. I was wondering is it possible to detect turning movements. Turning usually happens on the z axis. Is there a way to tell if body part has rotated 180 degrees with x and y coordinate.
I had few ideas like calculating the slope of the hand line. The slope tells us if the hand is tilted or not. If the slope is high or very low then the hand has rotated. Same concept for all other body parts. But I don't think that will work in all cases. Please check 2nd figure here https://cmu-perceptual-computing-lab.github.io/openpose/web/html/doc/md_doc_02_output.html to understand output of OpenPose.
I am trying to solve a seemingly easy problem related to device motion but couldn't figure out how to solve it. I have a situation where iPhone will move in a circle in the x-y plane. I need to find the angle between the iPhone's x and y axes relative to the center of rotation. The iPhone may be in portrait mode or landscape mode or in any angle in between relative to the line connecting iphone to the center of rotation. See the attached picture that explains the scenario.
The yaw change for a given rotation is the same regardless of this angle, so that doesn't really help. I am hoping that there would be some relationship that I can calculate for every small rotation and then find the best fit for the entire motion - but can't figure out that yet.
I appreciate any help or pointers.
(I am writing in pseudo-code since I don't know the API you are using, sorry.)
Here is how to get the axis and angle of the rotation.
Get the rotation matrices R1 and R2 at the beginning and end of your rotation directly from the API (see CMAttitude and CMRotationMatrix). Then, determine the angle and axis of the rotation R that brings R1 to align with R2. You get R as follows:
R = R1 * transpose(R2)
The angle of rotation R is
angle = acos((trace(R)-1)/2)
and its axis is
axis = { R[3][2]-R[2][3], R[1][3]-R[3][1], R[2][1]-R[1][2] }
For further details, please check Rotation matrix to axis angle and also Axis-angle.
I am not sure how to get the angle you are interested in. Nevertheless, I hope that the above helps.
Please don't use roll, pitch and yaw anything other than display. And don't try to integrate them, nothing good will come out.
Anyways, behind the rotation matrices there is integration. In other words, somebody already did the integration for you.
I want to use the iPhones's accelerometer to detect motions while driving. I'm a bit confused what the accelerometer actually measures, especially when driving a curve.
As you can see in the picture, a car driving a curve causes two forces. One is the centripetal force and one is the velocity. Imagine the iPhone is placed on the dashboard with +y-axis is pointing to the front, +x-axis to the right and +z-axis to the top.
My Question is now what acceleration will be measured when the car drives this curve. Will it measure g-force on the -x-axis or will the g-force appear on the +y axis?
Thanks for helping!
UPDATE!
For thoses interested, as one of the answers suggested it measures both. The accelerometer is effected by centrifugal force and velocity resulting in an acceleration vector that is a combination of these two.
I think it will measure both. But don't forget that the sensor will measure gravity as well. So when your car is not moving, you will still get accelerometer readings. A nice talk on sensors in smartphones http://www.youtube.com/watch?v=C7JQ7Rpwn2k&feature=results_main&playnext=1&list=PL29AD66D8C4372129 (it's on android, but the same type of sensors are used in iphone).
Accelerometer measures acceleration of resultant force applied to it (velocity is not a force by the way). In this case force is F = g + w + c i.e. vector sum of gravity, centrifugal force (reaction to steering centripetal force, points from the center of the turn) and car acceleration force (a force changing absolute value of instantaneous velocity, points along the velocity vector). Providing Z axis of accelerometer always points along the gravity vector (which is rare case for actual car) values of g, w and c accelerations can be accessed in Z, X and Y coordinates respectively.
Unless you are in free fall the g-force (gravity) is always measured. If I understand your setup correctly, the g-force will appear on the z axis, the axis that is vertical in the Earth frame of reference. I cannot tell whether it will be +z or -z, it is partly convention so you will have to check it for yourself.
UPDATE: If the car is also going up/downhill then you have to take the rotation into account. In other words, there are two frames of reference: the iPhone's frame of reference and the Earth frame of reference. If you would like to deal with this situation, then please ask a new question.
I noticed a really puzzling behavior on iPhone:
If I hold the phone in the vertical, and tilt it, the compass change.
I already figured the amount it changes is the same amount it would change for the same amount of tilting if it was in horizontal (ie: suppose that a vector coming from the screen is called Y, turning around Y does not matter the attitude of the iPhone results in a compass change).
I want to compensate that, my app was not made to you hold the phone in the horizontal (although I do plan also to allow some tilting in the X axis let's call it, from like 10 degrees to 135)
But I really could not figure how iPhone calculate the heading, thus where the heading vector actually points...
After some scientific style experiments, I found:
The iPhone has magnetometer, it has 3 axis, X, that goes from left to right from the screen. Y, that goes from bottom to up. And Z, that comes from behind the phone and comes to the front.
Earth magnetic field is as expected by the laws of physics not a sphere, in the location I am (brazil), it is slanted about 30 degrees. (meaning that I have to hold the phone in a 30 degrees angle to zero 2 axis).
One possible technique to calculate north, is use cross product of a vector tangential to the magnetic field (ie: the vector the magnetometer reports to you), and gravity. The result will be a vector that points east. If you wish you can make another cross product between east and gravity, resulting in a vector that points north.
Know that iPhone sensors are quite good, and every minor fluctuation and vibration is caught, thus it is good idea to use a lowpass filter, to remove the noise from the signal.
The iPhone itself, has a complex routine to determine the "true heading", I don't figured it completely, but it uses the accelerometer in some way to compensate for tilt. You can use the accelerometer and compensate back if that is your wish, for example if the phone is tilted 70 degrees, you can change the true heading by 70 degrees too, and the result will be the phone ignoring tilting.
Also the routine of true heading, verify if the iPhone is upside down or not. If we consider it in horizontal, in front of you as 0, then more or less at 135 degrees it decides that it is upside down, flipping the results.
Note the same coordinate system also apply to the accelerometer, allowing the use of vectors operations between accelerometer and magnetometer data without much fiddling.
I am trying to solve a tricky math problem, in a cocos2d for iphone context.
Basically I have a roullette wheel which is rotating over time.
I want to have a Sprite latch onto the wheel at certain points (like compass ordinal points N, S, E, W) and bounce off at all other points.
I have the image of the wheel rotating, and just need to solve the part where I can test for whether a sprite has intersected with the circle at the right point on the circle as it is rotating.
I think this question is going in the right direction, but I can't get my head around it. Can anyone help explain?
Best way to find a point on a circle closest to a given point
If I understand correctly:
First check the distance between the sprite and the centre of the roulette wheel. This will tell you if the sprite is at the edge of the wheel. (If not, nothing happens, right?)
Then, find the angle that the sprite makes from the "x-axis" of the roulette wheel.
spriteAngle = atan2(sprite.x - rouletteCentre.x, sprite.y - rouletteCentre.y)
You'll need to find the equivalent of the atan2() function. It usually returns an answer in radians; you may want to convert it to degrees or quarter-turns or something if you prefer.
Then, subtract the angle that the roulette wheel itself is rotated by (if the wheel itself is rotating, if not then you're already done). Make sure your angle measurement is consistent.
actualAngle = spriteAngle - rouletteRotationAngle
Note that actualAngle may be outside the range 0-360 degrees, and you will need to make it "wrap around".
Lastly, you will want to allow a small range of values as acceptable (e.g. 98 degrees to 102 might count as "North").
So you have a circle of radius r, with center (x0,y0).
A point lies outside of the circle, at coordinates (x,y). Your question is to find the closest point on the circle itself to the point (x,y).
The solution is simple. The closest projection of a point onto a circle is accomplished by a simple scaling. Thus,
d = sqrt((x-x0)^2 + (y-y0)^2)
xp = x0 + (x - x0)*r/d
yp = y0 + (y - y0)*r/d
The new point (xp,yp) will lie on the circle itself. To be honest, you would be better off to work in polar coordinates, with the origin at the center of the circle. Then everything gets much easier.
Your next question will be where did it hit on the circle? Don't forget the points of the compass on the circle are rotating with time. An atan2 function will give you the angle that the point (xp-x0,yp-y0) lies at. Most toolsets will have that functionality. See that I've subtracted off the origin here.