Can this still happen with my GPR Stabilizer??? Or does it cure it completly? I keep my dial at 5 so it's kinda stiff.

 

or just dont attempt em........

 

basically its physics based on gyroscope equastion that will best answer this

SF = mv'
Here, v is velocity, v' is the time rate of change of velocity (dv/dt) , and mv' is the time rate of change of linear momentum. If we chose a reference point O and r is a position vector to the particle, we can take the cross product of both sides of this equation to get an expression that relates the moment of the forces (Mo) acting on the particle to the angular momentum (Ho) of the particle with respect to the reference point O:


r x SF = r x mv'
Recognizing that r x SF is the Sum of the Moments ( SMo) of the forces acting on the particle about the reference point O, we can write:


SMo = r x mv'
Since the particle's angular momentum is Ho = r x mv, if we take the time derivative of angular momentum, we have:


d(Ho)/dt = d(r x mv)/dt
H'o = r' x mv + r x mv'
Recognizing that v = dr/dt = r' , we have:


H'o = r' x mr' + r x mv'
Since the cross product of two equivalent vectors is zero, r' x mr' = m(r' x r') = 0, therefore:


H'o = r x mv'
Substituting into the equation for the Sum of the Moments, we have:


SMo = H'o
That is, given a moving particle, the Sum of the Moments about a point O is equal to the time rate of change of the particle's angular momentum.

For a system of particles, we sum the moments of the forces of all the particles. In the following equation, Sf is the internal force acting on the ith particle due to all the other particles in the system:


Si[(r x SF)i + (r x Sf)i] = Si[H'o]i
The internal forces cancel out because corresponding pairs of internal forces are equal in magnitude and opposite in direction, therefore, Si(r x Sf)i = 0, and the resultant equation for a system of particles has the same form as the equation for a single particle:


Si[r x SF]i = Si[H'o]i
SMo = H'o
In words, this states that the Sum of the Moments about point O due to the external forces acting on a system of particles is equal to the time rate of change of the angular momentum of the system of particles about this same reference point O.

We recognize that any solid body is a system of particles, so this equation applies to the analysis of a gyroscope.

What we need now is an expression for the angular momentum Ho or its time derivative H'o that has attributes that we can physically measure such as mass, radius, angular velocity, and angular acceleration. If we consider a particle in the body having an incremental mass Dm and having an angular velocity w with respect to reference point O, since v = w x r, we can write:


[DHo]i = r x Dmivi
[DHo]i = [r x (w x r)]iDmi
Summing all the incremental angular momenta for all the particles in the body, we have:


Si[DHo]i = Si[r x (w x r)]iDmi
If we let D approach 0, then D[Ho]i and Dmi become differentials, and we can replace Si with integration. I will use z instead of the usual integral sign simply because I don't have the usual integral sign available:


zHodHo = zmr x (w x r) dm
Ho = zmr x (w x r) dm
If we place xyz reference coordinate axes at point O, we can define Ho, r, and w in terms of i, j, and k components as follows:


Ho = Hx i + Hy j + Hz k
r = x i + y j + z k
w = wx i + wy j + wz k
Substituting into the above integral expression for Ho, we have:


Hx i + Hy j + Hz k =
zm(x i + y j + z k) x [(wx i + wy j + wz k ) x (x i + y j + z k)]dm
Computing the cross products and combining terms gives:


Hx i + Hy j + Hz k =
[wxzm(y2+z2)dm - wyzmxy dm - wzzmxz dm] i
+ [- wxzmxy dm + wyzm(x2+z2)dm - wzzmyz dm] j
+ [- wxzmzx dm - wyzmyz dm + wzzm(x2+y2)dm] k
Recognizing that the above integrals are moments of inertia and products of inertia, we can write the above as the following scalar equations:


Hx = + Ixxwx - Ixywy - Ixzwz
Hy = - Iyxwx + Iyywy - Iyzwz
Hz = - Izxwx - Izywy + Izzwz
If

 

z=zZz [sm=sleeping.gif]

 

LMAO

 

k

 

I got a bad one at 50mph the other day comin off a wheelie. Slapped to left then right fast and hard. Gave it gas then pulled her straight. Just hard to think about giving it gas when its actually happening