Rotation and Thrust for Thinsats

The triangular version 2 thinsat has circular thrusters on each of the 3 corners, with areas of 16.86 cm2 and center radius of 9.27 cm.


Imagine a thinsat perched on a shallow cone, with a 120 degree apex, a face length of 1cm, and a diameter of cos(30°)*2π cm or 5.44cm. If we rolled a thinsat around the side of the cone, by rotating it 30 degrees, then adjusting the torque vector from the thrusters to lead the torque 90° (WAG) in front of the displacement. The thinsat "track" is 2π cm or 6.28cm, so after one turn around the cone the thinsat is ahead by 0.84 cm, or 48° ( = (1-cos(30°))*360° ). Straighten up the thinsat, and we've made a 48° turn. It takes around 200 seconds to turn 30° without stopping, and we will make on the order of 2π of those turns in sequence (WAG), so our 48° turn takes around 1200 seconds, or about 2 degrees per minute, or 120 degrees per hour. Not fast, but adequate for rolling a thinsat into a precise orientation.


A glancing collision, such as a micrometeoroid penetrating a thinsat near the edge, at a steep angle, could cause a thinsat to spin in a roll. How do we take that rotation out?

We can do this by adding torque at just under the rate of spin, opposing it. If the thinsat is wobbling, just a bit, as it spins, we can add a rotating torque with a component that opposes the angular momentum of the spinning disk, using the wobble to redistribute the tiny torque available from the rapidly switching electrochromic thrusters.

I don't have the math for this yet, so I will make some wild guesses. The force difference on a 5 cm diameter thruster (10% to 90% reflection adjustment range) is 1.6 times the light pressure (7.3μPa) times the thruster area (0.002 m2) or 14 nanonewtons. This is on a 10 cm lever arm, for a torque of 1.43 nN-m. As we rotate the torque vector around the surface of the disk, that is maximum at the 0°(one thruster) and 60° (two thruster) positions, as well as 120°, 180°, 240° and 270° positions. The force drops to 87% at the 30°, 90°, 150°, 210°, 270° and 330° positions. Overall, this averages to 95% torque, or 1.36e-9 N-m.

I still need to compute the moment of the thinsat, but it will probably will be around mr2/2 for a 10 cm radius, 5 gram disk disk, or 25 μkg-m2 . If it is set spinning at 5Hz, the angular frequency is 31.42 radians per second and the angular momentum is 6.3e-4 kg-m2/s.

Wild guessing: If we can rotate 10% of our 1.36e-9 N-m torque, or 1.36e-10 kg-m2/s2, into the spin axis, then the stopping time will be on the order of 6.7e-3 / 1.36e-10 or about 6 million seconds or about 70 days. That is a long time, and our damaged thinsat may need to be retired, but at least we can manuever it afterwards.

It is also possible that we can maneuver another sacrificial thinsat sideways into the spinning thinsat, and "move it into the spinning sawblade" to slow it down. That might be an interesting thing to try during prototype testing, but is probably not a good idea if there are other alternatives.

A third possibility is to add "roll thrusters" to the regular thrusters, with shallow gratings designed to reflect light at an angle and provide rotational optical thrust. If two of those each took up 10% of the thruster area, and deflected light by 30 degrees sideways (reducing front-back thrust by 0.5*0.2*(1-cos(30&deg)) or 1.3%), then the sideways thrust of each thruster would be 0.7nN, and the total torque for three thrusters would be 2e-10 Nm, reducing our stopping time to 27 days.

BETTER CALCULATIONS NEEDED! UPDATING FOR 5 grams and lower thrust needed! MoreLater

Rotation (last edited 2013-05-11 21:03:17 by KeithLofstrom)