Troubleshooting Interplanetary Cruise Calculations

In summary: Apparently, it was moved there. I have a post with the same title but a little bit more understanding. Hope this helps.No, what would HELP would be if you would define your variables and separate out your equations from the rest of the lines they are on so that things read more clearly (both in terms of meaning and in terms of typography)
  • #1
Viii
18
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Thread moved from the technical forums, so no Homework Template is shown.
Summary:: You have a spacecraft the size of a cube (2x2x2 m^3, m=3000 kg. It has 24 thrusters (2N each) and one main engine. For preparing your spacecraft for maneuvers, you want to rotate by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.
What am I doing wrong? Please read below.

I have this project assignment. I'll say the questions and explain what I did.

Q1) Minimum total maneuver to slew the spacecraft by 180 degrees=? (assume center of mass is in middle of cube).
Answer: I used this formula:
tb=sqrt(θm*Iz/n*F*L), where I already have θm=180 degrees or pi, Iz=(2*m*(b^2))/3=8000 kg*m^2, n=4, L=d/2=0.5 m, F=2N.
From all these I get tb=79.26 sec. So tm=4 (the thrusters)*tb=317.04 sec.

Q2) It is given a=0.95 and e=0.85. What is the temperature (in Celsius) when reaching Mars, when you know that only one of the faces is illuminated by the Sun and that the heat flow is distrubuted homogeneously?
Answer: I used this formula: T=(a/e)^(1/4) * ((S*An)/(σ*Atot))^(1/4).
Atot=24 m^2, right? So An=2 m^2. (A are surfaces). S=1.4 kW/m^2 and σ=5,67*10^(-8) W/m^2*K^4. These two are contstants.
With these numbers I found T=217.23 K, so T=-55.92 Celsius.

Q3) The power which is generated in the spacecraft can be considered as input power for the radiation balance. How much power (in Watt) should be dissipated inside the spacecraft , in order to achieve 0 Celsius?
Answer: 0 Celsius in Kelvin is 237.15, right? So I decided to use this formula: Pe=e*σ*T^4*Atot. I have everything, so I find Pe=3658.5 Watt.

I believe I'm doing something wrong with my calculations with the numbers, or I'm using wrong formulas. Whatever it is, the results above are not right.
What's you insight on this problem? Are the calculations wrong? Thanks in advance.
 
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  • #2
Viii said:
Summary:: You have a spacecraft the size of a cube (2x2x2 m^3, m=3000 kg. It has 24 thrusters (2N each) and one main engine. For preparing your spacecraft for maneuvers, you want to rotate by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.
What am I doing wrong? Please read below.

Minimum total maneuver to slew the spacecraft by 180 degrees=?
What does the question mean??
 
  • #3
I can't understand many of your variables, or the equations used, or the questions asked. The minimum maneuver would be the slowest rotation rate possible and then thrust to stop the rotation. How much time are you allowed for the rotation?
 
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  • #4
Dunno about most of it, but 0°C is 273 K, not 237.
 
  • #5
Viii said:
tb=sqrt(θm*Iz/n*F*L), where I already have θm=180 degrees or pi, Iz=(2*m*(b^2))/3=8000 kg*m^2, n=4, L=d/2=0.5 m, F=2N.
Let us try to figure out your variable names as @FactChecker would like. You should always do this for us. We should not need to guess.

It appears that you are using multi-character variable names with splats ("*") for multiplication rather than juxtaposition. Juxtaposition and single-character variable names would be more standard for handwritten or type-set physics work.

tb = time of burn.
lz = moment of inertia of a cube [about an axis passing through its edge !?]
n = number of thrusters firing
L = moment arm
F = thrust per thruster.

One error is obvious already. You didn't show your algebra, but it looks like a second error is lurking there. I am guessing that another error is in the assumption underlying the algebra which was not shown.

So how about you explain to us how you obtained the formula that you provided above.
 
  • #6
Hello everyone! I'm moving this to the homework forum, where it should be in the first place! I'd love to intearact with you there. Thank you all!
 
  • #7
Viii said:
Hello everyone! I'm moving this to the homework forum, where it should be in the first place! I'd love to intearact with you there. Thank you all!
Apparently, it was moved there. I have a post with the same title but a little bit more understanding. Hope this helps.
 
  • #8
Viii said:
Apparently, it was moved there. I have a post with the same title but a little bit more understanding. Hope this helps.
No, what would HELP would be if you would define your variables and separate out your equations from the rest of the lines they are on so that things read more clearly (both in terms of meaning and in terms of typography)
 
  • #9
Hi @Viii. As you have probably realized, what you have posted so far seems incomplete and impossible to understand, except by guesswork.

I suspect you haven't given us the complete question (word-for-word, and with any supplied diagrams). You may wish to fix this!
 
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  • #10
Viii said:
Hello everyone! I'm moving this to the homework forum, where it should be in the first place! I'd love to intearact with you there. Thank you all!
Viii said:
Apparently, it was moved there. I have a post with the same title but a little bit more understanding. Hope this helps.
No, your other/duplicate thread start was no more help, and was deleted and/or combined with this thread. Please use this thread here and do a great job of defining your problem statement and showing your work. Thanks.
 
  • #11
berkeman said:
No, your other/duplicate thread start was no more help, and was deleted and/or combined with this thread. Please use this thread here and do a great job of defining your problem statement and showing your work. Thanks.
Hello. I can't seem to find an edit button, so I can make my question more understanding. That's the reason I posted the other one (plus I didn't know it was moved here in homework threads). Thank you.
P. S: would it be better if I deleted this and wrote it again? Or no?
 
  • #12
Viii said:
Hello. I can't seem to find an edit button, so I can make my question more understanding. That's the reason I posted the other one (plus I didn't know it was moved here in homework threads). Thank you.
P. S: would it be better if I deleted this and wrote it again? Or no?
You need to explain your reasoning and show your work. The "other one" that you mention did not do either.

Personally, I am still waiting for you to respond to my post #5 from Thursday.
 
  • #13
Viii said:
would it be better if I deleted this and wrote it again? Or no?
No, multiple posting is not allowed. Just write a clarifying post now here in this thread, and please answer the questions you've been asked. Thank you.
 
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  • #14
Viii said:
Summary:: You have a spacecraft the size of a cube (2x2x2 m^3, m=3000 kg. It has 24 thrusters (2N each) and one main engine. For preparing your spacecraft for maneuvers, you want to rotate by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.
What am I doing wrong? Please read below.

I have this project assignment. I'll say the questions and explain what I did.

Q1) Minimum total maneuver to slew the spacecraft by 180 degrees=? (assume center of mass is in middle of cube).
Answer: I used this formula:
tb=sqrt(θm*Iz/n*F*L), where I already have θm=180 degrees or pi, Iz=(2*m*(b^2))/3=8000 kg*m^2, n=4, L=d/2=0.5 m, F=2N.
From all these I get tb=79.26 sec. So tm=4 (the thrusters)*tb=317.04 sec.

Q2) It is given a=0.95 and e=0.85. What is the temperature (in Celsius) when reaching Mars, when you know that only one of the faces is illuminated by the Sun and that the heat flow is distrubuted homogeneously?
Answer: I used this formula: T=(a/e)^(1/4) * ((S*An)/(σ*Atot))^(1/4).
Atot=24 m^2, right? So An=2 m^2. (A are surfaces). S=1.4 kW/m^2 and σ=5,67*10^(-8) W/m^2*K^4. These two are contstants.
With these numbers I found T=217.23 K, so T=-55.92 Celsius.

Q3) The power which is generated in the spacecraft can be considered as input power for the radiation balance. How much power (in Watt) should be dissipated inside the spacecraft , in order to achieve 0 Celsius?
Answer: 0 Celsius in Kelvin is 237.15, right? So I decided to use this formula: Pe=e*σ*T^4*Atot. I have everything, so I find Pe=3658.5 Watt.

I believe I'm doing something wrong with my calculations with the numbers, or I'm using wrong formulas. Whatever it is, the results above are not right.
What's you insight on this problem? Are the calculations wrong? Thanks in advance.
Rewritten problem:

Problem: You have a spacecraft the size of a cube (2x2x2 m3, m=3000 kg. It has 24 thrusters (2N each) and one main engine. The main engine is perpendicular to one of the faces of the cube and each has 4 attitude thrusters.
For preparing your spacecraft for maneuvers when reaching Mars, you want to rotate it by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.

Questions:
Q1) What is the minimum total maneuver to slew the spacecraft by 180 degrees? (Assume center of mass is in middle of the cube).
Q2) It is given a=0.95 and e=0.85. What is the temperature (in Celsius) when reaching Mars, when you know that only one of the faces is illuminated by the Sun and that the heat flow is distrubuted homogeneously?
Q3) The power which is generated in the spacecraft can be considered as input power for the radiation balance. How much power (in Watt) should be dissipated inside the spacecraft , in order to achieve 0 Celsius?

Note: All formulas and types used in the answers below are provided on slides by my professor.

Answers (what I calculated):
Q1 Answer: In order to find the minimum total maneuver, I firstly used this formula:
tb=sqrt(θmIz/nFL), where

tb = durationof burn (sec)
θm = minimum angle of rotation (rad)
Iz =moment of inertia around the rotational axis (kg m2)
n = number of thrusters used for the maneuver
F = thrust for each thruster (N)
L = perpendicular distance from the center of mass to the thrust vector (m)


I already have given from the problem that I want to rotate the spacecraft by 180 degrees. So I used θm=180 degrees=pi.
Iz is given by the formula Iz=(2m(b2))/3=8000 kg m2.
n=4 is given.
L=d/2=0.5 m
F=2N is given.

So, tb=sqrt(θmIz/nFL) = sqrt((pi*8000)/(4*2*0.5))=79.26 sec.

In order to find the minimum total maneuver time, I used this formula: tm=ntb, where
tm = minimunm time (sec)
n = number of thrusters used for the maneuver
tb = durationof burn (sec)


So, tm=ntb = 4*79.26=317.04 sec.

Q2 Answer:
In order to find the temperature, I used this formula: T=(a/e)(1/4) ((SAn)/(σAtot)(1/4) , where

T = temperature (K)
a = solar absorption (given in the question)
e = emissivity (given in the question)
S = solar constant (kW/m2) = 1.4 kW/m2 = 1400 W/m2
An = Surface perpendicular to the Sun (m2) = 2 m2 (right?)
σ = stefan-Boltzmann constant (W/m2K4) = 5.67 * 10-8 W/m2K4
Atot = total surface (m2)= 24 m2, right?


With these numbers I found T=(a/e)(1/4) ((SAn)/(σAtot)(1/4)= (0.95/0.85)(1/4)*((1400*2)/(5.67*10-8 * 24))(1/4). I found T=217.23 K, so T=-55.92 Celsius.

Q3 Answer: 0 Celsius in Kelvin is 273.15, right?

I used this formula: Pe=eσT4Atot, where


Pe =total amount of radiation emitted by an object per square meter (W/m2)
e = emissivity (given in Q2)
σ = stefan-Boltzmann constant (W/m2K4) = 5.67 * 10-8 W/m2K4
T = temperature (K) = 273.15 K
Atot = total surface (m2)= 24 m2


I have everything, so Pe=eσT4Atot=0.85*5.67 * 10-8*273.154*24. So, Pe=6438 Watt.

Professor said that my results are wrong. What is your opinion?

I personally believe I'm doing something wrong with my calculations with the numbers, or I'm using wrong formulas. Whatever it is, the results above are not right.
What's you insight on this problem? Are the calculations wrong? Thanks in advance.
 
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  • #15
hutchphd said:
What does the question mean??
I meant, What is the minimum total maneuver time. Please check the reply 'Rewritten problem' with orange letters in the answers. Thank you.
 
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  • #16
FactChecker said:
I can't understand many of your variables, or the equations used, or the questions asked. The minimum maneuver would be the slowest rotation rate possible and then thrust to stop the rotation. How much time are you allowed for the rotation?
The fact is that i don't know. That's why I'm calculating time of burn. Please check the reply 'Rewritten problem' with orange letters in the answers. It may help you more. Thank you.
 
  • #17
mjc123 said:
Dunno about most of it, but 0°C is 273 K, not 237.
You're right. Changed it. Thanks.
 
  • #18
jbriggs444 said:
Let us try to figure out your variable names as @FactChecker would like. You should always do this for us. We should not need to guess.

It appears that you are using multi-character variable names with splats ("*") for multiplication rather than juxtaposition. Juxtaposition and single-character variable names would be more standard for handwritten or type-set physics work.

tb = time of burn.
lz = moment of inertia of a cube [about an axis passing through its edge !?]
n = number of thrusters firing
L = moment arm
F = thrust per thruster.

One error is obvious already. You didn't show your algebra, but it looks like a second error is lurking there. I am guessing that another error is in the assumption underlying the algebra which was not shown.

So how about you explain to us how you obtained the formula that you provided above.
The formulas are given on slides by the professor. I have corrected the moltiplication process a bit. Please check the reply 'Rewritten problem' with orange letters in the answers. It may help you more. All the variables are there, and you are so right to remind me this. Thank you.
 
  • #19
phinds said:
No, what would HELP would be if you would define your variables and separate out your equations from the rest of the lines they are on so that things read more clearly (both in terms of meaning and in terms of typography)
Thank you for your feedback. You are right. Please check the reply 'Rewritten problem' with orange letters in the answers. It may help you more. Thank you.
 
  • #20
Steve4Physics said:
Hi @Viii. As you have probably realized, what you have posted so far seems incomplete and impossible to understand, except by guesswork.

I suspect you haven't given us the complete question (word-for-word, and with any supplied diagrams). You may wish to fix this!
You are so right. Thank you for saying this kidnly. Please check the reply 'Rewritten problem' with orange letters in the answers. It may help you more. Thank you.
 
  • #21
jbriggs444 said:
You need to explain your reasoning and show your work. The "other one" that you mention did not do either.

Personally, I am still waiting for you to respond to my post #5 from Thursday.
I'm so sorry. I answered above. Thank you.
 
  • #22
berkeman said:
No, multiple posting is not allowed. Just write a clarifying post now here in this thread, and please answer the questions you've been asked. Thank you.
I just did. Thank you for your patience.
 
  • #23
Viii said:
The fact is that i don't know. That's why I'm calculating time of burn. Please check the reply 'Rewritten problem' with orange letters in the answers. It may help you more. Thank you.
From the rewritten problem: "Q1) What is the minimum total maneuver to slew the spacecraft by 180 degrees?"
Answer: The minimal total maneuver to slew 180 degrees is 180 degrees.
Please be careful in your statement of the problem and do not force us to figure out your calculations just to know what the real question is.
 
  • #24
Your rewritten problem in Post #14, still has many issues which make it very hard to understand. Let me try to point-out/explain the issues I can easily spot.

My comments are in square brackets and bold:

Problem: You have a spacecraft the size of a cube (2x2x2 m3, m=3000 kg. It has 24 thrusters (2N each) and one main engine.

[Does this mean there are 4 thrusters on each face of the cube? What direction are they pointing? Do they operate as 2 pairs on each face, to give clockwise or anticlockwise motion about the face’s principle axis? Or maybe their directions can be adjusted.]

The main engine is perpendicular to one of the faces of the cube and each has 4 attitude thrusters.

[Are the ‘attitude thrusters’ the same as the ‘24 thrusters’ previously mentioned?]

For preparing your spacecraft for maneuvers when reaching Mars, you want to rotate it by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.

[I don’t understand which axis the 180 degree rotation is about, which 4 thrusters are used and which directions they fire.]

[Seem like (a) diagram(s) is/are missing?]


Questions:
Q1) What is the minimum total maneuver to slew the spacecraft by 180 degrees? (Assume center of mass is in middle of the cube).

[I guess you have missed out the key word ‘time’!]

Q2) It is given a=0.95 and e=0.85.

[I have no idea what a and e are, though I guess they could relate to absorbtivity and emissivity – in which case shouldn’t they be equal?}

What is the temperature (in Celsius) when reaching Mars, when you know that only one of the faces is illuminated by the Sun and that the heat flow is distrubuted homogeneously?

[Temperature of what? Front face? Back face? Other? Does the cube have any internal heat supply?]

Q3) The power which is generated in the spacecraft can be considered as input power for the radiation balance. How much power (in Watt) should be dissipated inside the spacecraft , in order to achieve 0 Celsius?

[Aha. That suggests Q2 wants you consider the case were internally generated power = 0. Though Q2 doesn’t state this. And what part of the spacecraft should be at 0 ºC? Maybe the centre?]

Note: All formulas and types used in the answers below are provided on slides by my professor.

Answers (what I calculated):
Q1 Answer: In order to find the minimum total maneuver, I firstly used this formula:
tb=sqrt(θmIz/nFL), where

[Formula has appeared by magic!]

tb = durationof burn (sec)
θm = minimum angle of rotation (rad)
[Why minimum? Does it include angular acceleration for half the time and angular deceleration for the other half of the time?]
Iz =moment of inertia around the rotational axis (kg m2)
n = number of thrusters used for the maneuver
F = thrust for each thruster (N)
L = perpendicular distance from the center of mass to the thrust vector (m)

I already have given from the problem that I want to rotate the spacecraft by 180 degrees. So I used θm=180 degrees=pi.
Iz is given by the formula Iz=(2m(b2))/3=8000 kg m2.

[What is ‘b’? What is the axis of rotation?]

n=4 is given.
L=d/2=0.5 m

[The cube is 2m on each side, so it is not clear what d means and what distance 0.5m means.]

[I’ll stop there, as I guess you might now see the problems we face trying to help! Have you given the complete, original, word-for-word problem?]
 
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  • #25
Viii said:
Iz is given by the formula Iz=(2m(b2))/3=8000 kg m2.
The moment of inertia of a cube depends crucially about what axis you choose.

Which axis do you think is relevant here? Why?

Viii said:
tb=sqrt(θmIz/nFL), where
This result looks like it derives from one of the SUVAT equations for rotational motion. I am skeptical, however, because it appears to be missing a factor of two. You have not justified this formula in any way.

So let us try to fill in with the justification that you should have provided...

The way to minimize the time needed to obtain a 180 degree rotation is to turn the thrusters on and leave them on until the desired rotation angle has been achieved. We assume that the craft is starting without rotation.

We will accept that the craft will be rotating rapidly at the end of the rotational burn. This is a questionable assumption. If one is trying to maneuver a craft, one would not normally want the craft to remain rotating.

We can start with ##\theta=\frac{1}{2}\alpha t^2## where ##\theta## is the rotation angle (in radians) achieved after ##t## seconds with constant angular acceleration ##\alpha##.

We want to solve for the required time ##t##. So we divide both sides by ##\frac{1}{2}\alpha## and get ##\frac{2 \theta}{\alpha}=t^2##.

Then we flip left for right and take the square root of both sides yielding ##t=\sqrt{\frac{2 \theta}{\alpha}}##

What remains is finding ##\alpha## in terms of the moment of inertia of the cube ##I_z##, the magnitude of the forces providing the torques ##F##, the moment arm for the torques ##L## and the number of forces ##n##. The result is that ##\alpha=\frac{nFL}{I_z}##.

[Your choice of variable name ##\theta_m## for the rotation angle was poor. Presented in plainish ASCII, ##\theta_m## is shown as θm which looks like angle ##\theta## multiplied by mass ##m##. I've chosen to dispense with the m subscript on ##\theta##].

In any case, once we substitute in for ##\alpha## in ##t=\sqrt{\frac{2 \theta}{\alpha}}##, we get $$t=\sqrt{\frac{2 \theta I_z}{nFL}}$$This does not match your proposed formula: tb=sqrt(θmIz/nFL). As predicted, the latter is missing a factor of two.
 
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  • #26
jbriggs444 said:
In any case, once we substitute in for ##\alpha## in ##t=\sqrt{\frac{2 \theta}{\alpha}}##, we get $$t=\sqrt{\frac{2 \theta I_z}{nFL}}$$This does not match your proposed formula: tb=sqrt(θmIz/nFL). As predicted, the latter is missing a factor of two.
We are given conflicting information:

a) the cube is 2m x 2m x 2m and L is the moment (lever) arm for the thrust; I would have thought L = 2m/2 = 1m [edit: or perhaps √2 m] with respect to the centre of mass;

b) L = d/2 (no idea what ‘d’ is) = 0.5m

I would guess that the missing factor of two is, at least in part, due to incorrectly interpreting the meaning of ‘L’ somewhere,
 
  • #27
Viii said:
In order to find the minimum total maneuver time, I used this formula: tm=ntb, where
tm = minimunm time (sec)
n = number of thrusters used for the maneuver
tb = durationof burn (sec)


So, tm=ntb = ...
So you've calculated the time (##t_b##) required for a single burn of four thrusters to obtain a rotation of 180 degrees. Then you've multiplied by four for some reason.

What is ##t_m## and why does it differ from ##t_b##?

If we fired four thrusters for ##t_b## seconds four times in a row, back to back, surely we would achieve a rotation angle of 180 degrees times sixteen?! -- Eight complete rotations.

180 degrees for the first burn ending with a rotation rate of ##\frac{360}{t_b}##
another 180 + 360 degrees for the second burn ending with a rotation rate of ##\frac{720}{t_b}##
another 180 + 720 degrees for the third burn ending with a rotation rate of ##\frac{1080}{t_b}##
another 180 + 1080 degrees for the fourth burn ending with a rotation rate of ##\frac{1440}{t_b}##

Add them up and that's a total of eight rotations (or just multiply half a rotation by four squared).
 
  • #28
Steve4Physics said:
[Does this mean there are 4 thrusters on each face of the cube? What direction are they pointing? Do they operate as 2 pairs on each face, to give clockwise or anticlockwise motion about the face’s principle axis? Or maybe their directions can be adjusted.]
There are eight corners on a cube. Each corner has three edges that meet there. If we had three thrusters at each corner and aligned them each to fire along an axis parallel to one of the edges, that'd give us 24 thrusters as specified.

That design would make the moment arms (about the center of gravity) equal to half the dimension of the cube (d = 2m, L=1m).

More realistically, one would likely optimize by arranging the three thrusters at each corner in a plane normal to a line from corner to center. Then the three thrusters on a corner would be at 120 degree angles to each other. The moment (lever) arm about the center of gravity would then be ##L=d\frac{\sqrt{3}}{2}=\sqrt{3}## meters. That's about a 70% improvement in performance.

If the professor has this latter interpretation in mind, he might be visualizing a rotation around a particular axis (an axis between the mid-points of two diagonally opposite and parallel edges). He would then assume the use of thrusters on the four corners not on the two selected edges. Finally, he might assume that one thruster on each corner is lined up exactly right for the intended rotation.

For a four-thruster burn, I believe that this choice of rotation axis and thruster selection optimizes the angular acceleration.
 
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  • #29
jbriggs444 said:
In any case, once we substitute in for ##\alpha## in ##t=\sqrt{\frac{2 \theta}{\alpha}}##, we get $$t=\sqrt{\frac{2 \theta I_z}{nFL}}$$This does not match your proposed formula: tb=sqrt(θmIz/nFL). As predicted, the latter is missing a factor of two.
This comes out right if you account for the fact that you need time to decelerate back to not-rotating.
 
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  • #30
willem2 said:
This comes out right if you account for the fact that you need time to decelerate back to not-rotating.
But then you would need to multiply by 2 to get the total burn time.

Either way, of course, we are left guessing at the reasoning used by @Viii since he hasn't told us what he was thinking.

We do have a clue though. When calculating the result, he uses ##\theta_m = \pi## (180 degrees). Not 90 degrees. And he does not multiply by two. He multiplies by n instead for some reason.
 
  • #31
FactChecker said:
From the rewritten problem: "Q1) What is the minimum total maneuver to slew the spacecraft by 180 degrees?"
Answer: The minimal total maneuver to slew 180 degrees is 180 degrees.
Please be careful in your statement of the problem and do not force us to figure out your calculations just to know what the real question is.
You're right. I wanted to say 'the minimum total maneuver time'. Thanks for pointing that out.
 
  • #32
Viii said:
You're right. I wanted to say 'the minimum total maneuver time'. Thanks for pointing that out.
What made me wonder about minimum time and ask about time available is that it seems as though in interplanetary flight there are months and even years available to get reoriented. A minimum fuel burn would be a tiny burn to get a slow rotation that would give 180 degrees after the total flight time and then a tiny burn to stop that slow rotation.
 
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  • #33
Steve4Physics said:
Your rewritten problem in Post #14, still has many issues which make it very hard to understand. Let me try to point-out/explain the issues I can easily spot.

My comments are in square brackets and bold:

Problem: You have a spacecraft the size of a cube (2x2x2 m3, m=3000 kg. It has 24 thrusters (2N each) and one main engine.

[Does this mean there are 4 thrusters on each face of the cube? What direction are they pointing? Do they operate as 2 pairs on each face, to give clockwise or anticlockwise motion about the face’s principle axis? Or maybe their directions can be adjusted.]

The main engine is perpendicular to one of the faces of the cube and each has 4 attitude thrusters.

[Are the ‘attitude thrusters’ the same as the ‘24 thrusters’ previously mentioned?]

For preparing your spacecraft for maneuvers when reaching Mars, you want to rotate it by 180 degrees. For the rotation, you'll use n=4 attitude thrusters per burn.

[I don’t understand which axis the 180 degree rotation is about, which 4 thrusters are used and which directions they fire.]

[Seem like (a) diagram(s) is/are missing?]


Questions:
Q1) What is the minimum total maneuver to slew the spacecraft by 180 degrees? (Assume center of mass is in middle of the cube).

[I guess you have missed out the key word ‘time’!]

Q2) It is given a=0.95 and e=0.85.

[I have no idea what a and e are, though I guess they could relate to absorbtivity and emissivity – in which case shouldn’t they be equal?}

What is the temperature (in Celsius) when reaching Mars, when you know that only one of the faces is illuminated by the Sun and that the heat flow is distrubuted homogeneously?

[Temperature of what? Front face? Back face? Other? Does the cube have any internal heat supply?]

Q3) The power which is generated in the spacecraft can be considered as input power for the radiation balance. How much power (in Watt) should be dissipated inside the spacecraft , in order to achieve 0 Celsius?

[Aha. That suggests Q2 wants you consider the case were internally generated power = 0. Though Q2 doesn’t state this. And what part of the spacecraft should be at 0 ºC? Maybe the centre?]

Note: All formulas and types used in the answers below are provided on slides by my professor.

Answers (what I calculated):
Q1 Answer: In order to find the minimum total maneuver, I firstly used this formula:
tb=sqrt(θmIz/nFL), where

[Formula has appeared by magic!]

tb = durationof burn (sec)
θm = minimum angle of rotation (rad)
[Why minimum? Does it include angular acceleration for half the time and angular deceleration for the other half of the time?]
Iz =moment of inertia around the rotational axis (kg m2)
n = number of thrusters used for the maneuver
F = thrust for each thruster (N)
L = perpendicular distance from the center of mass to the thrust vector (m)

I already have given from the problem that I want to rotate the spacecraft by 180 degrees. So I used θm=180 degrees=pi.
Iz is given by the formula Iz=(2m(b2))/3=8000 kg m2.

[What is ‘b’? What is the axis of rotation?]

n=4 is given.
L=d/2=0.5 m

[The cube is 2m on each side, so it is not clear what d means and what distance 0.5m means.]

[I’ll stop there, as I guess you might now see the problems we face trying to help! Have you given the complete, original, word-for-word problem?]
Answering your questions:
[Does this mean there are 4 thrusters on each face of the cube? What direction are they pointing? Do they operate as 2 pairs on each face, to give clockwise or anticlockwise motion about the face’s principle axis? Or maybe their directions can be adjusted.]

Imagine a cube. Each face has 4 thrusters, yes. They are pointing to the opposite direction of the face (see attached file). I think they operate as 1 pair on each face. But I believe that their directions can be adjusted.

[Are the ‘attitude thrusters’ the same as the ‘24 thrusters’ previously mentioned?]
Yes. They are the same. 24 in total. 4 in each face.

[I don’t understand which axis the 180 degree rotation is about, which 4 thrusters are used and which directions they fire.][Seem like (a) diagram(s) is/are missing?]
Please check the attached file. It's a drawing I did of the shape on the exercise paper. It doesn't state which axis, so I assumed and took the z axis. I also added one similar exercise the professor had in the past (see second attachment please).

[I guess you have missed out the key word ‘time’!]
Yes, I unfortunately did. I'm sorry.

[I have no idea what a and e are, though I guess they could relate to absorbtivity and emissivity – in which case shouldn’t they be equal?}
It is explained in the answers of that post. But to answer and your question here, yes, the are absorbtivity and emissivity. I asked the professor today and he said that they are absorbtivity anf emissivity of a specific material, called black epoxy, which the spacecraft is covered with. (Just to say that he didn't mention this when he gave us the problem).

[Temperature of what? Front face? Back face? Other? Does the cube have any internal heat supply?]
Temperature of the spacecraft , which is covered with black epoxy. I guess we choose which face. No mention for any internal heat supply.

[Aha. That suggests Q2 wants you consider the case were internally generated power = 0. Though Q2 doesn’t state this. And what part of the spacecraft should be at 0 ºC? Maybe the centre?]
Probably something like that. There is nowhere stated what oart should be at 0 Celsius, so is it stupid to assume the entire spacecraft temperature?😅

[Formula has appeared by magic!]
Somehow. It was given. You can see it also at the second attached file for another exercise, for a 90 degree maneuver.

[What is ‘b’? What is the axis of rotation?]
b is the side of the cube, which is 2. I assumed and took z and that it passes through it's egde.

[The cube is 2m on each side, so it is not clear what d means and what distance 0.5m means.]
d is half the side distance, which is the distance from the center of mass to the trust vector. So d=1. I assumed this also. It's nowhere given.

[I’ll stop there, as I guess you might now see the problems we face trying to help! Have you given the complete, original, word-for-word problem?]
Thank you for your patience. I have given the problem, but I'm so confused now myself. :oldcry:
 

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  • #34
jbriggs444 said:
The moment of inertia of a cube depends crucially about what axis you choose.

Which axis do you think is relevant here? Why?This result looks like it derives from one of the SUVAT equations for rotational motion. I am skeptical, however, because it appears to be missing a factor of two. You have not justified this formula in any way.

So let us try to fill in with the justification that you should have provided...

The way to minimize the time needed to obtain a 180 degree rotation is to turn the thrusters on and leave them on until the desired rotation angle has been achieved. We assume that the craft is starting without rotation.

We will accept that the craft will be rotating rapidly at the end of the rotational burn. This is a questionable assumption. If one is trying to maneuver a craft, one would not normally want the craft to remain rotating.

We can start with ##\theta=\frac{1}{2}\alpha t^2## where ##\theta## is the rotation angle (in radians) achieved after ##t## seconds with constant angular acceleration ##\alpha##.

We want to solve for the required time ##t##. So we divide both sides by ##\frac{1}{2}\alpha## and get ##\frac{2 \theta}{\alpha}=t^2##.

Then we flip left for right and take the square root of both sides yielding ##t=\sqrt{\frac{2 \theta}{\alpha}}##

What remains is finding ##\alpha## in terms of the moment of inertia of the cube ##I_z##, the magnitude of the forces providing the torques ##F##, the moment arm for the torques ##L## and the number of forces ##n##. The result is that ##\alpha=\frac{nFL}{I_z}##.

[Your choice of variable name ##\theta_m## for the rotation angle was poor. Presented in plainish ASCII, ##\theta_m## is shown as θm which looks like angle ##\theta## multiplied by mass ##m##. I've chosen to dispense with the m subscript on ##\theta##].

In any case, once we substitute in for ##\alpha## in ##t=\sqrt{\frac{2 \theta}{\alpha}}##, we get $$t=\sqrt{\frac{2 \theta I_z}{nFL}}$$This does not match your proposed formula: tb=sqrt(θmIz/nFL). As predicted, the latter is missing a factor of two.
Hello. I assumed the z axis. I chose it randomly, since it's not stated anywhere which one shall I take. By the way, love your answer. It's very detailed.

I took the same formula as in the picture attached (it's a similar problem). But now that you made the analysis it makes more sense. It was 90 degrees there, so pi/2, so the 2 factor was probably gone by that. Since now we have 180 degrees, we need the 2 factor. Right?
 

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  • #35
Steve4Physics said:
We are given conflicting information:

a) the cube is 2m x 2m x 2m and L is the moment (lever) arm for the thrust; I would have thought L = 2m/2 = 1m [edit: or perhaps √2 m] with respect to the centre of mass;

b) L = d/2 (no idea what ‘d’ is) = 0.5m

I would guess that the missing factor of two is, at least in part, due to incorrectly interpreting the meaning of ‘L’ somewhere,
I considered d=1m so L=d/2=0.5m. Bad mistake.:headbang:
 

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