Will the duration of traveling to Ceres using the same tech developed for going to Mars be proportional to...
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I've read somewhere (I dont have right now the source and I dont know if it's trustworthy) that something that would take a significant amount of time for going to Mars would be the acceleration and deceleration process. If this is the case, a travel to a larger distance using the same technology could be not proportional to the distance . Is this so? Will the duration of traveling to Ceres using the same technology developed for going to Mars be proportional to the distance employed for going to Mars or not? If it isnt proportional what would it be, shorter or larger?
spacex mars ceres
asked 18 hours ago
PabloPablo
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add a comment |
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I've read somewhere (I dont have right now the source and I dont know if it's trustworthy) that something that would take a significant amount of time for going to Mars would be the acceleration and deceleration process. If this is the case, a travel to a larger distance using the same technology could be not proportional to the distance . Is this so? Will the duration of traveling to Ceres using the same technology developed for going to Mars be proportional to the distance employed for going to Mars or not? If it isnt proportional what would it be, shorter or larger?
spacex mars ceres
asked 18 hours ago
PabloPablo
1525
New contributor
Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I've read somewhere (I dont have right now the source and I dont know if it's trustworthy) that something that would take a significant amount of time for going to Mars would be the acceleration and deceleration process. If this is the case, a travel to a larger distance using the same technology could be not proportional to the distance . Is this so? Will the duration of traveling to Ceres using the same technology developed for going to Mars be proportional to the distance employed for going to Mars or not? If it isnt proportional what would it be, shorter or larger?
spacex mars ceres
asked 18 hours ago
PabloPablo
1525
New contributor
Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've read somewhere (I dont have right now the source and I dont know if it's trustworthy) that something that would take a significant amount of time for going to Mars would be the acceleration and deceleration process. If this is the case, a travel to a larger distance using the same technology could be not proportional to the distance . Is this so? Will the duration of traveling to Ceres using the same technology developed for going to Mars be proportional to the distance employed for going to Mars or not? If it isnt proportional what would it be, shorter or larger?
spacex mars ceres
spacex mars ceres
asked 18 hours ago
PabloPablo
1525
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Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
PabloPablo
1525
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Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
PabloPablo
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asked 18 hours ago
PabloPablo
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asked 18 hours ago
PabloPablo
1525
1525
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2 Answers
2
active
oldest
votes
$begingroup$
Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity.
To go from Earth to Ceres, you need to initially travel faster than you do to get to Mars, otherwise you will not be going fast enough to get that far away from the Sun before the Sun's gravity slows you down and starts pulling you in.
There are lots of possible times to depart and arrive, each of which needs a different amount of rocket propulsion. These are conveniently gathered in something called a "pork chop plot". You can calculate them online
For instance for Mars you might get:
This shows departure dates on the x axis, flight times on the y axis and the fuel needed (for the departure from Earth at least, but you can aerobrake on arrival) by the colour.
For Ceres you get:
So the most fuel efficient journeys need almost twice as much fuel (more because you can't aerobrake) and take about twice as long.
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
$endgroup$
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
|
show 1 more comment
$begingroup$
In addition to Steve Linton's excellent answer there's a simple pattern:
To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory anyway. (Not to say that you just pile on the engines, the normal approach is to use planetary flybys to gain or shed velocity. Nothing has been launched to a target beyond Jupiter without using Jupiter for a boost.)
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
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add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity.
To go from Earth to Ceres, you need to initially travel faster than you do to get to Mars, otherwise you will not be going fast enough to get that far away from the Sun before the Sun's gravity slows you down and starts pulling you in.
There are lots of possible times to depart and arrive, each of which needs a different amount of rocket propulsion. These are conveniently gathered in something called a "pork chop plot". You can calculate them online
For instance for Mars you might get:
This shows departure dates on the x axis, flight times on the y axis and the fuel needed (for the departure from Earth at least, but you can aerobrake on arrival) by the colour.
For Ceres you get:
So the most fuel efficient journeys need almost twice as much fuel (more because you can't aerobrake) and take about twice as long.
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
$endgroup$
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
|
show 1 more comment
$begingroup$
Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity.
To go from Earth to Ceres, you need to initially travel faster than you do to get to Mars, otherwise you will not be going fast enough to get that far away from the Sun before the Sun's gravity slows you down and starts pulling you in.
There are lots of possible times to depart and arrive, each of which needs a different amount of rocket propulsion. These are conveniently gathered in something called a "pork chop plot". You can calculate them online
For instance for Mars you might get:
This shows departure dates on the x axis, flight times on the y axis and the fuel needed (for the departure from Earth at least, but you can aerobrake on arrival) by the colour.
For Ceres you get:
So the most fuel efficient journeys need almost twice as much fuel (more because you can't aerobrake) and take about twice as long.
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
$endgroup$
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
|
show 1 more comment
$begingroup$
Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity.
To go from Earth to Ceres, you need to initially travel faster than you do to get to Mars, otherwise you will not be going fast enough to get that far away from the Sun before the Sun's gravity slows you down and starts pulling you in.
There are lots of possible times to depart and arrive, each of which needs a different amount of rocket propulsion. These are conveniently gathered in something called a "pork chop plot". You can calculate them online
For instance for Mars you might get:
This shows departure dates on the x axis, flight times on the y axis and the fuel needed (for the departure from Earth at least, but you can aerobrake on arrival) by the colour.
For Ceres you get:
So the most fuel efficient journeys need almost twice as much fuel (more because you can't aerobrake) and take about twice as long.
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
$endgroup$
Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity.
To go from Earth to Ceres, you need to initially travel faster than you do to get to Mars, otherwise you will not be going fast enough to get that far away from the Sun before the Sun's gravity slows you down and starts pulling you in.
There are lots of possible times to depart and arrive, each of which needs a different amount of rocket propulsion. These are conveniently gathered in something called a "pork chop plot". You can calculate them online
For instance for Mars you might get:
This shows departure dates on the x axis, flight times on the y axis and the fuel needed (for the departure from Earth at least, but you can aerobrake on arrival) by the colour.
For Ceres you get:
So the most fuel efficient journeys need almost twice as much fuel (more because you can't aerobrake) and take about twice as long.
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
answered 17 hours ago
Steve LintonSteve Linton
7,76512044
7,76512044
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
|
show 1 more comment
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
5
5
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
14 hours ago
4
4
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
12 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
6 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
$begingroup$
@corsiKa, for an optimal trajectory, time of flight and dV are (mostly) unrelated. If you're going for a suboptimal trajectory, a shorter time of flight requires a greater dV because you're accelerating more at the start, and then you need to get rid of that excess speed when you arrive. A longer time of flight also requires a greater dV because you're deliberately going on a less-direct path than the optimal route (eg. if you're going to Mars, you'd go further out from the Sun than Mars's orbit, then come back in from the direction of the asteroid belt.)
$endgroup$
– Mark
5 hours ago
1
1
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
@corsiKa: More delta V is required for Ceres because Ceres is farther from the sun, so we need to add more energy to the orbit to get there. Because Ceres is farther from the sun, its orbit has a longer period. As Loren Pechtel says, the optimal flight time is (about) half the slower orbital period, so the flight time is longer.
$endgroup$
– Ross Millikan
5 hours ago
|
show 1 more comment
$begingroup$
In addition to Steve Linton's excellent answer there's a simple pattern:
To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory anyway. (Not to say that you just pile on the engines, the normal approach is to use planetary flybys to gain or shed velocity. Nothing has been launched to a target beyond Jupiter without using Jupiter for a boost.)
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
$endgroup$
add a comment |
$begingroup$
In addition to Steve Linton's excellent answer there's a simple pattern:
To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory anyway. (Not to say that you just pile on the engines, the normal approach is to use planetary flybys to gain or shed velocity. Nothing has been launched to a target beyond Jupiter without using Jupiter for a boost.)
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
$endgroup$
add a comment |
$begingroup$
In addition to Steve Linton's excellent answer there's a simple pattern:
To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory anyway. (Not to say that you just pile on the engines, the normal approach is to use planetary flybys to gain or shed velocity. Nothing has been launched to a target beyond Jupiter without using Jupiter for a boost.)
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
$endgroup$
In addition to Steve Linton's excellent answer there's a simple pattern:
To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory anyway. (Not to say that you just pile on the engines, the normal approach is to use planetary flybys to gain or shed velocity. Nothing has been launched to a target beyond Jupiter without using Jupiter for a boost.)
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
answered 9 hours ago
Loren PechtelLoren Pechtel
5,9411021
5,9411021
add a comment |
add a comment |
Pablo is a new contributor. Be nice, and check out our Code of Conduct.
Pablo is a new contributor. Be nice, and check out our Code of Conduct.
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