Spacecraft Observation Part 2
We continue the conversation . Suppose we want to look at something 20 cm in size on the surface of the moon. We take a passing load, but not more than 100 kg. From the point of landing, it is assumed that he still has to get under his own power to the desired orbit with the required flow rate of the total impulse 800 m / s. To fly to the moon with an overhang of 100 kg, it is sometimes enough just to add fuel to the acceleration stage. But then the device maneuvers itself, and the designer of the device will not rely on us. The fact that we are flying together, it may become clear when our companion will be ready.
Given:
Mo = 100 kg
dV = 800 m / s
dLm = 0.2 m
The solution to the problem can be found here . I did not find, therefore ...
1. Getting to the place
This engine is offered on the forum.
P = 13.3 N - thrust
I = 2688 m / s - specific impulse
M do = 0.55 kg - engine weight
according to the Tsiolkovsky formula M1 = M0 * exp (-dV / I) - the mass remaining after the maneuver
M1 = 74.25 kg
')
It is interesting to calculate the engine running time t = Mt / m = (Mo-M1) / (P / I) = 5200 s = 1.45 hours
Here I would like to return to the old dispute why more powerful engines are needed . The fact is that in 1.45 hours our apparatus could fly around the moon. There are a lot of difficulties: the amount of fuel will increase, it is necessary to stabilize the device for an hour and a half.
Without going into details, suppose we are satisfied with the engine operation time of 150 s.
Then the required engine thrust is P = m * I = (Mo-M1) * I / 150 = 576 N
This is about 11D458M , it has a slightly better impulse I = 2963, but the mass is more Mdu = 3 kg and the length is half a meter (for the time being we neglect).
M1 = 76.33 kg
So we have left M1 = Mpn + Md + Mskhrt
Where
Mpn - payload mass
MSHRT - the mass of the storage and supply system of rocket fuel (tanks, pumps, valves) is approximately 15 kg for fuel weighing 24 kg.
Mpn = 58 kg
2. Fly, take pictures, send
2.1 Target hardware
Optics (not an expert therefore excerpts from the forum)
Conditions: height of 10 km, resolution 10cm / pix
The matrix from the AFA Z / I DMC II 250 is 17216 x 14656 pixels in size, with a pixel size of 5.6 microns, the matrix with a physical size of approximately 96.5x82 mm
At the output we have: Focal length 560 mm, angular field of 13 degrees.
The shutter speed should not exceed 1/7500.
The hardest lens, while in the amount of just Mop.
Information transmission equipment. Here the whole question is whether we will be able to gain access to the antennas of remote space communications. Since from a distance of 400,000 km the signal attenuates strongly. Either this is the mass and power of the onboard transmission system, or 70 m plate on Earth. Perhaps someone will tell in the comments.
2.2 Construction
Normal untight frame. Approximately 1 * 0.5 * 0.5 m. Solar batteries are supposed to be placed on the surface without orientation, which is not very successful, but more on that later.
Mk = Kk * Mo - where Kk is the mass coefficient of the structure
2.3 On-board power engineering
When placing the panels on the edges of the device (the ends are occupied by the engine and lens), it means that at best we will have 1/4 of the area of ​​the batteries oriented on the Sun, and they will weigh everything.
Take XB = 1.5 kg / m2
Then Msb = 1.5 * 1 * 0.5 * 4 = 3 kg
Nmax = 200W, if found with an efficiency of 30%
The real average daily power, taking into account half the time spent in the shadow of the Moon, will be approximately 1500 watts.
Of these, about a third must be stored in batteries.
As = 0,005 kg / W
Mac = 2.5 kg
Thermal mode is needed optics and processors BVS. I think enough radiators and heating with electricity.
Integrated propulsion system. In addition to the specified brake motor, we will need 6 thruster (2 per axle) for orientation and stabilization of the apparatus. The calculation in the next part.
2.4 BVS and other electronics
Now it's pretty simple. You can almost put the arduino. The main thing to pay attention to is radiation resistance. Let it be MBVs = 2kg with relays, wires and radiators.
2.5 Motion Control System
First of all, you need to determine the direction to the center of the moon - this is what we will shoot. It is possible to shoot at a slight inclination, but optimally strictly downwards. On the spacecraft of observation, as a rule, an infrared plotter of the local vertical (ICMV) is installed. The device is relatively small, weighs one and a half kilograms. Perhaps there is easier. It would be good for two to reserve.
Speed ​​increment meter to determine when to turn off the engine.
Angular velocity sensors are needed to determine the oscillation of the apparatus and the corresponding correction using a power gyro. In the simplest case, it is a pancake on the rotor of the motor. It works like this: we accelerate the motor, the device turns in one direction, we brake - in the other. When the maximum rotational speed is reached, for braking the rotor, the apparatus is held by the engines.
It would be nice to know where we are. This is either a star or solar sensor. More reliably stellar, since the sun is harder to catch. And at the same time they would support another domestic startup .
Total for all sensors need 4-5 pounds. Power gyroscopes 5 more.
On this round out. We will continue to consider and even can build graphs.
Given:
Mo = 100 kg
dV = 800 m / s
dLm = 0.2 m
The solution to the problem can be found here . I did not find, therefore ...
1. Getting to the place
This engine is offered on the forum.
P = 13.3 N - thrust
I = 2688 m / s - specific impulse
M do = 0.55 kg - engine weight
according to the Tsiolkovsky formula M1 = M0 * exp (-dV / I) - the mass remaining after the maneuver
M1 = 74.25 kg
')
It is interesting to calculate the engine running time t = Mt / m = (Mo-M1) / (P / I) = 5200 s = 1.45 hours
Here I would like to return to the old dispute why more powerful engines are needed . The fact is that in 1.45 hours our apparatus could fly around the moon. There are a lot of difficulties: the amount of fuel will increase, it is necessary to stabilize the device for an hour and a half.
Without going into details, suppose we are satisfied with the engine operation time of 150 s.
Then the required engine thrust is P = m * I = (Mo-M1) * I / 150 = 576 N
This is about 11D458M , it has a slightly better impulse I = 2963, but the mass is more Mdu = 3 kg and the length is half a meter (for the time being we neglect).
M1 = 76.33 kg
So we have left M1 = Mpn + Md + Mskhrt
Where
Mpn - payload mass
MSHRT - the mass of the storage and supply system of rocket fuel (tanks, pumps, valves) is approximately 15 kg for fuel weighing 24 kg.
Mpn = 58 kg
2. Fly, take pictures, send
2.1 Target hardware
Optics (not an expert therefore excerpts from the forum)
Conditions: height of 10 km, resolution 10cm / pix
The matrix from the AFA Z / I DMC II 250 is 17216 x 14656 pixels in size, with a pixel size of 5.6 microns, the matrix with a physical size of approximately 96.5x82 mm
At the output we have: Focal length 560 mm, angular field of 13 degrees.
The shutter speed should not exceed 1/7500.
The hardest lens, while in the amount of just Mop.
Information transmission equipment. Here the whole question is whether we will be able to gain access to the antennas of remote space communications. Since from a distance of 400,000 km the signal attenuates strongly. Either this is the mass and power of the onboard transmission system, or 70 m plate on Earth. Perhaps someone will tell in the comments.
2.2 Construction
Normal untight frame. Approximately 1 * 0.5 * 0.5 m. Solar batteries are supposed to be placed on the surface without orientation, which is not very successful, but more on that later.
Mk = Kk * Mo - where Kk is the mass coefficient of the structure
2.3 On-board power engineering
When placing the panels on the edges of the device (the ends are occupied by the engine and lens), it means that at best we will have 1/4 of the area of ​​the batteries oriented on the Sun, and they will weigh everything.
Take XB = 1.5 kg / m2
Then Msb = 1.5 * 1 * 0.5 * 4 = 3 kg
Nmax = 200W, if found with an efficiency of 30%
The real average daily power, taking into account half the time spent in the shadow of the Moon, will be approximately 1500 watts.
Of these, about a third must be stored in batteries.
As = 0,005 kg / W
Mac = 2.5 kg
Thermal mode is needed optics and processors BVS. I think enough radiators and heating with electricity.
Integrated propulsion system. In addition to the specified brake motor, we will need 6 thruster (2 per axle) for orientation and stabilization of the apparatus. The calculation in the next part.
2.4 BVS and other electronics
Now it's pretty simple. You can almost put the arduino. The main thing to pay attention to is radiation resistance. Let it be MBVs = 2kg with relays, wires and radiators.
2.5 Motion Control System
First of all, you need to determine the direction to the center of the moon - this is what we will shoot. It is possible to shoot at a slight inclination, but optimally strictly downwards. On the spacecraft of observation, as a rule, an infrared plotter of the local vertical (ICMV) is installed. The device is relatively small, weighs one and a half kilograms. Perhaps there is easier. It would be good for two to reserve.
Speed ​​increment meter to determine when to turn off the engine.
Angular velocity sensors are needed to determine the oscillation of the apparatus and the corresponding correction using a power gyro. In the simplest case, it is a pancake on the rotor of the motor. It works like this: we accelerate the motor, the device turns in one direction, we brake - in the other. When the maximum rotational speed is reached, for braking the rotor, the apparatus is held by the engines.
It would be nice to know where we are. This is either a star or solar sensor. More reliably stellar, since the sun is harder to catch. And at the same time they would support another domestic startup .
Total for all sensors need 4-5 pounds. Power gyroscopes 5 more.
On this round out. We will continue to consider and even can build graphs.
Source: https://habr.com/ru/post/367847/
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