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 don't have right now the source and I don't 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 isn't proportional what would it be, shorter or longer?
spacex mars ceres
edited Feb 24 at 0:23
Glorfindel
2091210
asked Feb 22 at 12:53
PabloPablo
267110
$endgroup$
add a comment |
$begingroup$
I've read somewhere (I don't have right now the source and I don't 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 isn't proportional what would it be, shorter or longer?
spacex mars ceres
edited Feb 24 at 0:23
Glorfindel
2091210
asked Feb 22 at 12:53
PabloPablo
267110
$endgroup$
add a comment |
$begingroup$
I've read somewhere (I don't have right now the source and I don't 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 isn't proportional what would it be, shorter or longer?
spacex mars ceres
edited Feb 24 at 0:23
Glorfindel
2091210
asked Feb 22 at 12:53
PabloPablo
267110
$endgroup$
I've read somewhere (I don't have right now the source and I don't 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 isn't proportional what would it be, shorter or longer?
spacex mars ceres
spacex mars ceres
edited Feb 24 at 0:23
Glorfindel
2091210
asked Feb 22 at 12:53
PabloPablo
267110
edited Feb 24 at 0:23
Glorfindel
2091210
asked Feb 22 at 12:53
PabloPablo
267110
edited Feb 24 at 0:23
Glorfindel
2091210
edited Feb 24 at 0:23
Glorfindel
2091210
edited Feb 24 at 0:23
Glorfindel
2091210
2091210
asked Feb 22 at 12:53
PabloPablo
267110
asked Feb 22 at 12:53
PabloPablo
267110
asked Feb 22 at 12:53
PabloPablo
267110
267110
add a comment |
add a comment |
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 delta-V (more because you can't aerobrake) and take about twice as long.
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
$endgroup$
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
2
$begingroup$
@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
Feb 23 at 1:11
|
show 3 more comments
$begingroup$
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 Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
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add a comment |
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2 Answers
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2 Answers
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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 delta-V (more because you can't aerobrake) and take about twice as long.
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
$endgroup$
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
2
$begingroup$
@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
Feb 23 at 1:11
|
show 3 more comments
$begingroup$
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 delta-V (more because you can't aerobrake) and take about twice as long.
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
$endgroup$
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
2
$begingroup$
@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
Feb 23 at 1:11
|
show 3 more comments
$begingroup$
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 delta-V (more because you can't aerobrake) and take about twice as long.
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
$endgroup$
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 delta-V (more because you can't aerobrake) and take about twice as long.
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
edited Feb 23 at 17:43
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
answered Feb 22 at 13:20
Steve LintonSteve Linton
9,00812449
9,00812449
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
2
$begingroup$
@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
Feb 23 at 1:11
|
show 3 more comments
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
2
$begingroup$
@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
Feb 23 at 1:11
6
6
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
$begingroup$
DV is delta v, I take it?
$endgroup$
– Acccumulation
Feb 22 at 16:41
5
5
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Yes. Note that delta V is not directly proportional to fuel; in a staged rocket the relationship is roughly exponential.
$endgroup$
– Skyler
Feb 22 at 18:12
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
Like I'm 5, can you explain why a longer flight time requires a higher dV?
$endgroup$
– corsiKa
Feb 23 at 0:54
$begingroup$
@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
Feb 23 at 1:08
$begingroup$
@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
Feb 23 at 1:08
2
2
$begingroup$
@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
Feb 23 at 1:11
$begingroup$
@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
Feb 23 at 1:11
|
show 3 more comments
$begingroup$
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 Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
$endgroup$
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 Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
$endgroup$
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 Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
$endgroup$
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 Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
answered Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
answered Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
answered Feb 22 at 21:36
Loren PechtelLoren Pechtel
6,1611221
6,1611221
add a comment |
add a comment |
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