SES100_Fall2023_Project1_Part4_FlightSystem
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Arizona State University *
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Course
100
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
6
Uploaded by EarlBadger3914
SES 100: Introduction to Exploration
Project 1: Flight System Design (50 pts.)
Group Number: 6
Names of Group Members: AJ Azevedo, Ryan Hofmann, Colton
Skinner, Alex Winkler, Christian Hoffman, Dylan Prince
Instructions
Please answer all the questions posed in this assignment and upload your solutions as
a pdf. Feel free to scan any handwritten calculations and notes but include everything in
one pdf. All group members should work together on the assignment—everyone should
submit an identical copy of the group’s solutions. Remember to consult the “General
Guidelines” document for basic information about the missions and camera options.
Please show your work for any calculations. When you perform calculations, please
remember to keep track of and report your units—and respect significant figures.
Part 1: Command & Data Handling (12 pts.)
What is the data volume for a single image, in bits and bytes?
The camera is 6000 by 8000 pixels, meaning there is 48000000 Pixels
We want it in color so each pixel is 24 bits or 3 bytes
meaning there is 144000000 bytes or 1152000000 bits
How many images will you collect during your imaging traverse?
We will try to collect at least 10 images during the imaging traverse
What is the total science data volume required for your imaging traverse?
1,440,000,000 bytes or 11,520,000,000 bits
What is the interval of time between each image?
If we are moving at a rate of 65m/s and since the area we are covering is 5m/pixel it
means that we are covering 240000000 m^2 but since we are moving horizontally we
only are covering 30000 meters length wise per picture. In order to get a good layout of
Venus we want our images to align with each other, so 30000/65= 461.52 seconds
between each image in order to get the perfect alignment of images.
What is the rate at which the spacecraft computer must record data from the camera (in
units of bits per second, bps)?
If our total data volume is 11,520,000,000 bits and we take 1 photo every 461.52
seconds with a total of 10 photos, we divide 11,520,000,000 by 4615.2 which gives up
2,496,099.84 bps
Part 2: Telecommunications (10 pts.)
Assume that your relay orbiter can communicate with the Deep Space Network (DSN)
on Earth for 8 hours per day. What data rate is required to send all your science data
back to Earth in 90 days (e.g., in units of bps or kbps)?
data rate = bits/time
time = 90 days*8 hours *3600 seconds = 2,592,000 seconds
data rate = x bits/2,592,000 seconds = bps
data rate = 144,000,000/2,592,000 = 444.444 bps
The data rate required to send all the science data back to earth in 90 days would be
444.444 bps
Using the approximate scaling below, how large of an antenna on the relay orbiter
would be required to achieve the required data rate?
???? ????
=
10 𝑘???
3 𝐴𝑈
?𝑖?????? ?? ????ℎ
(
)
2
𝐴?????? ?𝑖??????
1 ?
(
)
2
1m*[(data rate)/(10kbps*(3au/0.76au)^2]^½
1m*[(444.444 bps)/(10000bps*(3.95)^2]^½
1m*[(444.444 bps)/(156,025 bps)]^½
0.0534m
5.34cm
The antenna on the relay orbiter would have to be 5.34 cm
Part 3: Attitude, Determination, and Control System (ADCS) (10 pts.)
For your mission, what is the iFOV of your camera?
The iFOV of our camera is 75
μrad.
The performance of an ADCS is defined in terms of how precisely it can point the
camera (pointing control), how accurately it knows where the camera is pointed
(pointing knowledge), and how much the camera moves or shakes over a given time
(pointing stability). Pointing control and knowledge are both angular quantities (e.g., with
units of arcseconds) while pointing stability is a maximum angular deviation per time
interval (e.g., with units of arcseconds per second).
Assume an exposure time of 50 milliseconds per image.
Please define and justify requirements for your ADCS:
●
Pointing knowledge
The requirements for our ADCS for pointing knowledge should be around 5 by 4
degrees.
0.2*6000*75 = 90000/1000000 = 0.09 rad = 5.15 degrees
0.2*4000*75 = 60000/1000000 = 0.06 rad = 3.43 degrees
●
Pointing control
The requirements for our ADCS for pointing knowledge should be around 5 by 4
degrees.
●
Pointing stability
The pointing stability needed for our ADCS should be around 36.75 μrad in 50ms.
0.49*75 = 36.75
Part 4: Motion Blur (15 pts.)
Motion blur is a major issue in planetary imaging. The problem is particularly severe for
flyby missions (e.g., New Horizons at Pluto) where the relative velocity between the
camera and the object is >>1 km/s. Motion blur is orders-of-magnitude less problematic
for a mission involving atmospheric flight—but is still worth considering.
Write an equation for the amount of motion blur (
b
, units of meters) in a single pixel as a
function of the wind speed (
v
, units of meters per second) and the exposure time (
t
,
units of seconds).
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? = 𝑣?
At the altitude of 47 km on venus the average wind speed is
≅
55 m/s. Our exposure time using
the standard lens would be 1/32 seconds.
b
= 55?/? * (1/32)?
1.72m
? =
For your specific mission, what is the motion blur in fractions of a pixel? For example, if
the motion blur is 10 m and the pixel scale (i.e., spatial resolution of individual pixels) is
100 m, then the motion blur could be described as 0.1 pixels.
? = 1. 72/3. 53 = 0. 49
The motion blur in fractions of a pixel is 0.49
The conventional requirement for planetary geology is motion blur <0.5 pixels. Does
your mission meet this requirement?
Our mission does meet the requirement for motion blur as 0.49 is less than the
maximum 0.5 pixels.
Your mission could incorporate a scan platform to eliminate motion blur entirely. Write
an equation for the angular speed at which the camera would need to rotate to eliminate
motion blur (ω, in units of radians per second) as a function of the wind speed (v), the
exposure time (
t
), and the balloon altitude (
d
, in units of meters). Solve the equation for
your mission. How fast would the scan platform need to rotate the camera?
d=47,000m
v=55m/s
t=1/32s
???
−1
(1. 72/47000)
=
0. 00209°
(
)*(pi/180) =
0. 00209
3. 66 * 10
−5
ω = (3. 66 * 10
−5
) * 32 = 1. 17 * 10
−3
???/?
The camera would need to rotate at 0.07°/s to completely counteract the motion blur
Part 5: Contributions (3 pts.)
Please briefly describe what each group member contributed to the assignment.
●
AJ Azevedo and Ryan Hofmann did calculations and a drawing for part 4
●
Dylan Prince did calculations and work on part 1
●
Colton Skinner did work on part 1
●
Christian Hoffman did calculations for part 2
●
Alex Winkler did work on part 1 and part 2 and cleaned up answers and Part 3
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