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posted by martyb on Sunday April 11 2021, @10:50PM   Printer-friendly
from the good-things-come-to-those-who-wait dept.

Straight from NASA we have word of a delay in the first flight of Ingenuity on Mars.

Mars Helicopter Flight Delayed To No Earlier Than April 14 - Nasa Mars:

Based on data from the Ingenuity Mars helicopter that arrived late Friday night, NASA has chosen to reschedule the Ingenuity Mars Helicopter's first experimental flight to no earlier than April 14.

During a high-speed spin test of the rotors on Friday, the command sequence controlling the test ended early due to a "watchdog" timer expiration. This occurred as it was trying to transition the flight computer from 'Pre-Flight' to 'Flight' mode. The helicopter is safe and healthy and communicated its full telemetry set to Earth.

The watchdog timer oversees the command sequence and alerts the system to any potential issues. It helps the system stay safe by not proceeding if an issue is observed and worked as planned.

The helicopter team is reviewing telemetry to diagnose and understand the issue. Following that, they will reschedule the full-speed test.

NASA has a web site devoted to Ingenuity.


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  • (Score: 2, Interesting) by Anonymous Coward on Monday April 12 2021, @04:10AM (3 children)

    by Anonymous Coward on Monday April 12 2021, @04:10AM (#1136249)

    Why are you taking your area ratios as the ratio of the lengths? What are the rotor widths?

    The lift equation of the rotors should go as the air density times the velocity squared of the rotor times the area, so if you are taking ratios, you need to ratio those. I'll accept your pressure ratio, but not your area ratio. And if you look at v-squared, the ratio of that will go as the square of the ratio of the radiuses, so (1200 / 480)^2 = 6.25. So now to make them equal, we need 0.006 x 6.25 x (A1/A2) =1, so A1/A2 needs to be about 26. But, as you noted, the lengths are 2.5 times longer on the Mars RC, now we're down to the widths of the rotors need to be about 10x wider.

    Remarkably, can't find specifics on the rotor design. Just lots of fluff pieces, but you can have pretty wide blades on those counter rotating designs, so I'm not sure what your big objection is.

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  • (Score: 0) by Anonymous Coward on Monday April 12 2021, @12:54PM (2 children)

    by Anonymous Coward on Monday April 12 2021, @12:54PM (#1136343)

    Is it the area of the actual blades that matters or the area that is covered by the rotor disk?

    This uses the rotor disk. No blade is going to have area of 50 m^2:

    The planform area of the rotor disk is 50 meters squared (A).

    https://sciencing.com/calculate-lift-rotor-blades-7680704.html [sciencing.com]

    • (Score: 0) by Anonymous Coward on Monday April 12 2021, @05:49PM (1 child)

      by Anonymous Coward on Monday April 12 2021, @05:49PM (#1136553)

      Rotor disk area is what matters for any rotary blade system, be it helicopter or windmill. For fixed wings it is wingspan×airspeed.

      • (Score: 0) by Anonymous Coward on Monday April 12 2021, @07:02PM

        by Anonymous Coward on Monday April 12 2021, @07:02PM (#1136602)

        Does that refer to the circle traced out by the blades as they spin? Or the actual physical surface area of the blades.

        If the latter, how can it be 50 m^2 for a normal helicopter?