Tag Info

New answers tagged

1

Depending on the shape of the universe the luminosity distance is given by : \begin{equation} d_L(z) = \left\{ \begin{array}{rl} \frac{(1 + z) c}{H_0 \sqrt{|\Omega_k|}} \sin \left[ \sqrt{|\Omega_k|} \int _0 ^z \frac{dz'}{H(z')/H_0} \right] & \mbox{for $k = 1$} \\ \frac{(1 + z) c}{H_0} \int _0 ^z \frac{dz'}{H(z')/H_0} & \mbox{for $k = 0$} \\ ...


2

Did you mean 12.8 billion light years away? If so, in this case the distance was estimated by measuring a rough spectrum for the GRB. The NASA article I've linked says: In certain colors, the brightness of a distant object shows a characteristic drop caused by intervening gas clouds. The farther away the object is, the longer the wavelength where this ...


0

For relatively close objects, the distance can be measured through measurements of parallax (http://en.wikipedia.org/wiki/Parallax ).


3

The accelerometer measures the negative of gravity plus any upwards acceleration see NOTE#1 $$ acc = -(g+\ddot{x}) $$ and you want to integrate $\ddot{x}$ to get speed $v=\dot{x}$ and position $x$. So your expressions should be $$v(t)=-\int_0^t ( acc+g)\,{\rm d}t \\ x(t)=-\int_0^t \int_0^t ( acc+g)\,{\rm d}t\,{\rm d}t$$ You also know that the final ...


2

Like Ross said, you need to try to remove the overall constant part of your acceleration data. Most accelerators report gravity all of the time. So even if your phone was sitting still on a desk, it would report and acceleration of -9.8 m/s^2 "down", however down lined up with the axes of your phone. Looking at your data a bit, for the first 1/3 of your ...


2

If your accelerations are in $g$, you should multiply the second term in column C by $9.8$. The velocities will then be in m/sec and your integration to make D will be in meters. You may then want to apply any constraints you know of. If you know the velocity starts and ends at zero, you might want to add or subtract a constant acceleration to the data to ...



Top 50 recent answers are included