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AAE450 Spring 2009 Final Sizing and Trajectory Design for 100 g/10kg Payloads [Levi Brown] [Mission Ops] March 12, 2009 1.

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Presentation on theme: "AAE450 Spring 2009 Final Sizing and Trajectory Design for 100 g/10kg Payloads [Levi Brown] [Mission Ops] March 12, 2009 1."— Presentation transcript:

1 AAE450 Spring 2009 Final Sizing and Trajectory Design for 100 g/10kg Payloads [Levi Brown] [Mission Ops] March 12, 2009 1

2 AAE450 Spring 2009 [Levi Brown] [Mission Ops] Final Sizing and Trajectory Determination 2 10 kg Payload LLO Payload Mass (kg) 185.0 Thrust (mN)75 mdot (mg/s)4.4 Mismatch Position (km)949 ΔV (m/s)470 Additional Propellant Mass (kg) 20.5 Results Initial Mass (kg)490.8 Cost (Million $)3.6 Time of Flight (days) 290 100 g Payload LLO Payload Mass (kg) 154.9 Thrust (mN)75 mdot (mg/s)4.8 Mismatch Position (km)6786 ΔV (m/s)382 Additional Propellant Mass (kg) 18.5 Results Initial Mass (kg)415.5 Cost (Million $)3.2 Time of Flight (days) 250 Note: Cost only includes launch, power, and propellant cost These numbers will change based on the lander freeze Wed night, but the change is expected to be small

3 AAE450 Spring 2009 [Levi Brown] [Mission Ops] 3 If we don’t match spirals, how eccentric are we? Depends on location of intersection 1.Close to Earth -- ΔV low but have not escaped Earth yet -- Spiral back to Earth 2.Close to Moon -- ΔV high -- Comparable eccentricity -- Several months to circularize Conclusion: 1. Our “matching spiral” model is the way to go 2. Better solutions with more sophisticated model -- Reduced ΔV -- Closer to Moon and not as eccentric -- Introduce coasting and variable thrust to nearly eliminate ΔV 3. System is capable of achieving mission

4 AAE450 Spring 2009 Back-up Slides [Levi Brown] [Mission Ops] 4

5 AAE450 Spring 2009 [Levi Brown] [Mission Ops] 5 Plot of Trajectory for 10 kg Payload NOTE: 100 g trajectory similar

6 AAE450 Spring 2009 [Levi Brown] [Mission Ops] 6 Plot of 10 kg Trajectory with Inertial Velocity Direction at End Point NOTE: 100 g trajectory similar

7 AAE450 Spring 2009 [Levi Brown] [Mission Ops] 7 More Notes This analysis for a 400 km LEO to a 25 km LLO 100 g: Requires 1.18 kW 10 kg: Requires 1.26 kW Initial eccentricity0.73 Time to circularize to 25 km ≈ 105 days Propellant Cost≈ 20 kg Circularization Much better than 4 years as previously estimated Spacecraft mass low in this analysis=more time Nearly identical additional propellant cost Approx. 40% increase in flight time Introduces more failure modes Less accurate model Comparison to Matching Spiral Model


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