I am for a not ball earth.

I was trying to put it in a way to get you to think about it.

as I get it, the earth is spining at 831.9KM a hour?

from the link it says 0.4 percent differences!

that seems to small?

Things don't spin at a speed - that's the wrong unit of measure. Things spin at angular rates, like degrees per second, or perhaps revolutions per unit of time, like rpm. Earth spins at 15 degrees per hour, which is barely perceptible. Yes, at the equator that equates to a fast speed, but that means little on its own when calculating the implications for the change in weight.

The equation we want for this is the one describing centripetal force and circular motion : a = v

^{2} / r

Earth's equatorial radius is around 6371km, or 6,371,000m

So our 'v' at the equator is the circumference of the equator (the distance) divided by the time it takes to travel that distance - one day. 24 hours is 86400 seconds, so v = 6371000 * pi * 2 / 86400 = 463 m/s, or 1667 km/h, which is about 1000mph.

Then we plug in our v and r figures into the equation and we get a = 463

^{2} / 6371000 = 0.0336 m/s

^{2}So, if our typical 'g' is 9.81 m/s

^{2}, then our figure at the equator will be 9.81 - 0.0336 = 9.776

So, not really much of a difference, although measurable with reasonable accurate scales, and significant if you are trying to achieve a world record in an event of some kind. It's worth bearing in mind though that other factors influence gravity itself around the globe, so the exact figure won't be this simple.