### Sidebar-1: Exponentials

Previously, we had decided to trade precision for convenience and model fuel consumption in an overly simplistic way.

BACKGROUND. Our notional spacecraft gains propulsion by accelerating one gram per second to .71c (change water particles to ions and then accelerate these ions to 71% light speed) to gain thrust by ejecting these ions from spacecraft. We do this process in an even and consistent matter for entire quantity of water ions throughout the entire second to increase velocity of spaceship 10 m/s. If we consistently do this throughout the entire flight; then, we've achieved g-force for that spaceship. (Upcoming sidebar: Specific Impulse). Theory of relativity tells us that the one gram of orginal mass will grow to 1.42 grams when it reaches 71% ligt speed and ejects from spacecraft. Conservation of momemtum tells us that 1.42 grams of relativistic mass times .71c will propel a 30 mT spaceship's mass in the opposite direction to gain additional 10m/s in speed. Consuming one gram/sec throughout an entire day gives us 86,400 grams/day (= 86.4 kg/dy). Thus, the daily percentage of spaceship's mass converted to kinetic energy = (86.4kgms/30,000 kgms) /day = .288%/dy (=D% for rest of sidebar).

Finally, we traded convenience for precision by treating this D% quantity in a linear way instead of the more precise exponential way. For example, to determine total fuel requirement for a ten day flight, we would multiply .29%/day by 10 days to get 2.9% and then multiple that value times ship's initial mass (also known as TOGW). Comparing to current technology, current aircraft require much more then 3% of Take Off Gross Weight to travel for a few hours of constant speed flight.

Math model: Fuel = D% * T * TOGW

This gives us a good approximation but it ignores inevitable system inefficiencies as well as decreasing gross weight (GW) throughout the flight. We'll continue to ignore inefficiencies for this sidebar, but we'll concentrate on the fact that decreasing GW means decreasing fuel consumption throughout powered flight.

Before we use exponentials, let's construct a table where we determine fuel requirements for each day of a ten day spaceflight.

Above table indicates that as fuel consumes without replacement then overall gross weight (GW) of spaceship decreases. Thus, while conservation of momentum requires same percentage of expended mass to maintain same velocity increase; overall quantity of fuel consumption rate will decrease. (This phenomena is observed all the time in current times.) Thus, a linear forecast of fuel consumption gives fuel planners a good approximation of fuel requirements, but it's slightly too much. (Good thing, too, cuz this slightly mitigates the much less then 100% efficiency that we've been assuming.)

Another thing, we need a better way to determine everdecreasing GW. Above table shows an elemental,but tedious way of doing this; construct a table to show every day's fuel consumption and resulting GW.

Better way: use simple exponential function. Everyday of flight, ship's mass decreases by .29%. Thus, we can determine 2nd day's GW by multiplying starting GW (recall TOGW) by 1 - .0029 = 0.9971. This gives us GW = 997.1 mT for start of 2nd day; to get GW at end of 2nd day (start of 3rd day), multiple .9971 times this value to get 994.2 mT (= 997.1 mt * .9971). After ten repetitions, we determine that GW at beginning of 10th day is 974.2mT. A quicker way would be the following arithmetic operation:

GWi = TOGW(1-D%)**i

GWi= gross weight at end of the ith day.

TOGW = Take Off Gross Weight = space ship's initial GW at beginning of trip.

D% = daily percentage of gross used for fuel consumption

i = number of days into mission.

GW10 = 1000mT (1-.0029)**10 = 971.4 mT

(NOTE: Slightly less then figure in 10th row of above table because that is the GW at the start of the 10th day.

To further analyze this discrepancy let's examine a 100 day flight at 10 day increments by using exponentials.

BACKGROUND. Our notional spacecraft gains propulsion by accelerating one gram per second to .71c (change water particles to ions and then accelerate these ions to 71% light speed) to gain thrust by ejecting these ions from spacecraft. We do this process in an even and consistent matter for entire quantity of water ions throughout the entire second to increase velocity of spaceship 10 m/s. If we consistently do this throughout the entire flight; then, we've achieved g-force for that spaceship. (Upcoming sidebar: Specific Impulse). Theory of relativity tells us that the one gram of orginal mass will grow to 1.42 grams when it reaches 71% ligt speed and ejects from spacecraft. Conservation of momemtum tells us that 1.42 grams of relativistic mass times .71c will propel a 30 mT spaceship's mass in the opposite direction to gain additional 10m/s in speed. Consuming one gram/sec throughout an entire day gives us 86,400 grams/day (= 86.4 kg/dy). Thus, the daily percentage of spaceship's mass converted to kinetic energy = (86.4kgms/30,000 kgms) /day = .288%/dy (=D% for rest of sidebar).

Finally, we traded convenience for precision by treating this D% quantity in a linear way instead of the more precise exponential way. For example, to determine total fuel requirement for a ten day flight, we would multiply .29%/day by 10 days to get 2.9% and then multiple that value times ship's initial mass (also known as TOGW). Comparing to current technology, current aircraft require much more then 3% of Take Off Gross Weight to travel for a few hours of constant speed flight.

Math model: Fuel = D% * T * TOGW

This gives us a good approximation but it ignores inevitable system inefficiencies as well as decreasing gross weight (GW) throughout the flight. We'll continue to ignore inefficiencies for this sidebar, but we'll concentrate on the fact that decreasing GW means decreasing fuel consumption throughout powered flight.

Before we use exponentials, let's construct a table where we determine fuel requirements for each day of a ten day spaceflight.

Table 1. First 10 Days. | |||||
---|---|---|---|---|---|

Time | StartGW | Rate | Fuel | Cumm | Cumm % |

Day | mT | %/day | mT | Fuel | |

0 | 1000.0 | 0.00% | 0 | 0 | 0.00% |

1 | 1000.0 | 0.29% | 2.900 | 2.90 | 0.29% |

2 | 997.1 | 0.29% | 2.892 | 5.79 | 0.58% |

3 | 994.2 | 0.29% | 2.883 | 8.67 | 0.87% |

4 | 991.3 | 0.29% | 2.875 | 11.55 | 1.15% |

5 | 988.5 | 0.29% | 2.867 | 14.42 | 1.44% |

6 | 985.6 | 0.29% | 2.858 | 17.27 | 1.73% |

7 | 982.7 | 0.29% | 2.850 | 20.12 | 2.01% |

8 | 979.9 | 0.29% | 2.842 | 22.97 | 2.30% |

9 | 977.0 | 0.29% | 2.833 | 25.80 | 2.58% |

10 | 974.2 | 0.29% | 2.825 | 28.62 | 2.86% |

Above table indicates that as fuel consumes without replacement then overall gross weight (GW) of spaceship decreases. Thus, while conservation of momentum requires same percentage of expended mass to maintain same velocity increase; overall quantity of fuel consumption rate will decrease. (This phenomena is observed all the time in current times.) Thus, a linear forecast of fuel consumption gives fuel planners a good approximation of fuel requirements, but it's slightly too much. (Good thing, too, cuz this slightly mitigates the much less then 100% efficiency that we've been assuming.)

Another thing, we need a better way to determine everdecreasing GW. Above table shows an elemental,but tedious way of doing this; construct a table to show every day's fuel consumption and resulting GW.

Better way: use simple exponential function. Everyday of flight, ship's mass decreases by .29%. Thus, we can determine 2nd day's GW by multiplying starting GW (recall TOGW) by 1 - .0029 = 0.9971. This gives us GW = 997.1 mT for start of 2nd day; to get GW at end of 2nd day (start of 3rd day), multiple .9971 times this value to get 994.2 mT (= 997.1 mt * .9971). After ten repetitions, we determine that GW at beginning of 10th day is 974.2mT. A quicker way would be the following arithmetic operation:

GWi = TOGW(1-D%)**i

GWi= gross weight at end of the ith day.

TOGW = Take Off Gross Weight = space ship's initial GW at beginning of trip.

D% = daily percentage of gross used for fuel consumption

i = number of days into mission.

GW10 = 1000mT (1-.0029)**10 = 971.4 mT

(NOTE: Slightly less then figure in 10th row of above table because that is the GW at the start of the 10th day.

To further analyze this discrepancy let's examine a 100 day flight at 10 day increments by using exponentials.

Table 2. Daily Rate = 0.29% of Ship's GW | |||||||
---|---|---|---|---|---|---|---|

Time (Days) | StartGW | Ave Rate | EndGW | CummFuel | Exp. | Linear | |

Start | End | mT | %/day | mT | mT | %TOGW | %TOGW |

0 | 10 | 1000 | 0.286% | 971.4 | 28.6 | 2.86% | 2.90% |

0 | 20 | 1000 | 0.282% | 943.6 | 56.4 | 5.64% | 5.80% |

0 | 30 | 1000 | 0.278% | 916.6 | 83.4 | 8.34% | 8.70% |

0 | 40 | 1000 | 0.274% | 890.3 | 109.7 | 10.97% | 11.60% |

0 | 50 | 1000 | 0.270% | 864.8 | 135.2 | 13.52% | 14.50% |

0 | 60 | 1000 | 0.267% | 840.1 | 159.9 | 15.99% | 17.40% |

0 | 70 | 1000 | 0.263% | 816.0 | 184.0 | 18.40% | 20.30% |

0 | 80 | 1000 | 0.259% | 792.7 | 207.3 | 20.73% | 23.20% |

0 | 90 | 1000 | 0.256% | 770.0 | 230.0 | 23.00% | 26.10% |

0 | 100 | 1000 | 0.252% | 747.9 | 252.0 | 25.21% | 29.00% |

NOTES for future sidebars:

Modify this side bar to use GWt = TOGW * e**-D%t

GWt = gross weight of spaceship after t days.

TOGW = TakeOff GW, initial weight of spaceship.

D% = daily loss of GW due to daily fuel consumption

t = length of flight in decimal days

Brief sidebar discussing revision of D% to reflect greater precision from even smaller time increments. Conclusion will likely be that the increased precision will be relatively slight.

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