Closed chaals closed 3 years ago
In example 9, the phone is described as being vertical, so accelerationIncludingGravity would register gravity affecting the y-axis, not the z-axis. This means the value should be { x: v^2/r, y: 9.8, z:0 }
accelerationIncludingGravity
{ x: v^2/r, y: 9.8, z:0 }
In example 9, the phone is described as being vertical, so
accelerationIncludingGravitywould register gravity affecting the y-axis, not the z-axis. This means the value should be{ x: v^2/r, y: 9.8, z:0 }