it still is the same in all directions seen from the ship on an about.

What does "on an about" mean?

]]>Or isn't that correct?

]]>Rick is right.

Only if you ignore that awful typo.

The thought is that the x wise coordinate is so much bigger compared to the meterstick on the ship with mass M

You're comparing a coordinate to a size? Perhaps you meant to compare the length in the x-axis to a size?

Even though the gravitational force seen from the ship is equal in all directions.

If there are a finite number of things in your system, gravitational force can not be the same in all directions.

So basically the gravityforce starts to move perpendicular to movement direction, the faster the ship goes.

Why?

]]>GmM√(1-(v/c)²)/r² = F(x)

Why call this F(x) if there is no F in the equation?

But what is F(y,z)?

This question doesn't seem to make sense.

The thought is that the x wise coordinate is so much bigger compared to the meterstick on the ship with mass M, Even though the gravitational force seen from the ship is equal in all directions.

So basically the gravityforce starts to move perpendicular to movement direction, the faster the ship goes.

I've been counting again: GmM/r^2 -> GmMsqrt(1 - (v/c)^2)/r^2 + -2GmM/r (for v -> c)

Can this be proven then?

]]>Rick is right. In F(2) where does the 2 go? Besides gravitomagnetism? I know of no transducer that you put in magnetism - electricity and get gravity, or vice versa.

]]>GmM√(1-(v/c)²)/r² = F(x)

Why call this F(x) if there is no F in the equation?

But what is F(y,z)?

This question doesn't seem to make sense.

]]>Make an assumption: Gravity becomes solely perpendicular at high velocities, cutting through the sun like paper from particleaccelerators

Can some of you on mathisfunforum callibrate the force instead?

GmM√(1-(v/c)²)/r² = F(x)

But what is F(y,z)?

Where G is gravity constant, v is speed, m & M are interacting masses and x,y & z are coordinates & y,z is the plane based on y,z coordinates.

]]>Oh heck, would you mind defining all of them?

]]>This is the gravitomagnetic force:

]]>