cubinator

Don't lick that.

The discovery of an alien civilization would be one of the most realistic drivers for a human push to the stars as well. If it's anything like KSP1, it'll be there for those who wish to pursue it, but won't take away from other experiences and challenges in the game.

5:20 or so, pulsar on the left monitor

Many of the locations within the US along that path are typically cloudy during that time of year, so that's another thing to consider.

They can drink ethanol and instead of breaking it down store it in a reservoir organ.

I know a handful of obscure Star Wars facts. For instance, that Jabba the Hutt reproduced by parthenogenesis.

If this was like 9 hours later it'd affect my airplane flight.

I'm not looking those orbital elements up, I feel it goes too far against the nature of the project. I am even using my own simplistic measurement of the Moon's distance which I know is slightly inaccurate but that I helped obtain myself. For the purpose of this experiment, I don't know the distance to the Sun or the length of Mars' orbit, or that Earth orbits the Sun. I know the Moon's size and distance, the Earth's axial tilt and the time of day and year, the Moon's period as observed since antiquity, and the coordinates of each observer. My assumptions are very simplistic, so as to show that we can perform science even with simple equipment. I can find the Moon's position and velocity in the sky using a mere few timestamps and math, and I find beauty in that. All of this could be bypassed by looking things up, but this is more of an art project than an actual experiment. NASA knows the distance to Mars better than I could ever estimate using my backyard telescope, but I am hopefully showing people that they have the power to achieve great things no matter what tools they have in front of them. I'm also showing a fundamental example of the scientific method, in which discoveries are made with an open mind and careful consideration of what can be observed from the world.

That's basically what I'm doing, the problem is I need the Moon's position and velocity rather exactly. Although I'm not to the point of doing numerical integration, since I already know where the few really important moments are. It should take care of the Earthly rotational motion problem, anyway. The unwritten step is that the Moon moves at 1 km/s, so every 1 km of muntain goes by in 1 second, making the units of km and s interchangeable in this setting. I have drawn it out in MATLAB and on paper, but if you're anything like me you'll probably spend a lot of time finding the nearest roughly spherical object in your house and rotating it around in your hand while staring at it. Bonus points if you shine lights at it. Let me find my drawing, I think I have a nice-ish one. Here it is: I have drawn in the Earth's axis, the Moon's velocity, and the three locations of observation. You can imagine the Moon moving across the face of Earth while Earth slowly rotates. This is what it looked like from Mars, except it was mostly dark because the Sun was shining on Earth from behind.

I can find roughly what the angle is based on the time spent behind the Moon and the known angular velocity and size of the Moon. I still need to account for Mars' angular velocity later, but it's enough to have a rough estimate of the amplitude of the effect of lunar mountains.

If Mars went behind the Moon at a 45 degree angle to the surface, and there was a 3 km dip between the first contact and second contact, then the additional time due to that dip would be 3*sqrt(2) = 4.25 seconds. If Mars went in at a shallow angle of 30 degrees, the effect of the dip would be 3 / cos(60*) = 6 seconds. Thus, a shallow entry angle amplifies the time difference due to lunar terrain greatly, enough to potentially account for a large part of the observed difference. A shallower angle also would cause Mars to pass over a longer stretch of the landscape, so shallower slopes would be able to make a difference as long as they have the altitude change overall.

I think that if it clipped something on the outskirts of Mare Orientale it could have gotten 3 km or so, I'm going to try to figure out if the angle Mars enters the Moon at amplifies that effect.

I calculated the velocity vector at Mars-set of Minnesota based on the vector between my position at t1 and t2, and the velocity vector at Mars-rise using the vector between my position at t3 and t4. Then I found the angle between the components of those vectors that are parallel to the YZ plane, because the Moon's motion is assumed to be 0 in the X direction (which is the direction from Earth pointing away from the sun). The angle of Minnesota's motion in that plane changed by 0.15769 radians, and the cosine of that angle is 0.9876. So, the difference in the observed velocity of the Moon due to Earth's rotation between the observations should be about 1.1%. The observed difference was about 20%. I'm not yet sure why you think this would cause the rise and set times to be different - I would think that Mars should come out from the Moon the same way it went in, because the angular velocity of the Moon and Mars are basically constant.

I did photograph Mars and the Moon before and after the event, so I can verify the topography if needed, but I have been seeing if I can get by without using photos so as to make the determination using even simpler data.

I figured that the Moon's angular velocity wouldn't change by as much as multiple percents in an hour or a few, as it only moved a couple degrees during the whole thing. I will see if Earth's rotation is enough to create such a difference - the observer is moving in a different direction after an hour, maybe that is not so insignificant.

Thanks for selecting my thread! I've been working on it on and off during my end-of-year travels. There's a couple things I'm still stumped on but I'm confident that the determination is possible, and I may even get more information than I anticipated.

Assuming the Moon's orbital period is 2360592 seconds, and that while Mars is rising and setting its own motion through the sky is negligible, the Moon's angular velocity is w = 2*pi/T, and Mars' angular size is w*time to rise or set. The rise and set times observed in Minnesota were different for an as-of-yet unknown cause, but they can be used nonetheless to estimate the angular size of Mars: Rise time estimate Set time estimate 21.4665 arcseconds 25.9684 arcseconds

I solved a Rubik's cube in 8.1 seconds.

I would be surprised if it had any effect, the Moon and Mars are both out of atmosphere so I'd think it'd affect them both equally. But science is full of surprises. The duration of transit is mostly dependent on how the Moon moves through space and how 'deeply' Mars goes behind the Moon from each vantage point. The timing is affected by the Earth's rotation and the motion of Moon and Mars.

I think I will also need to come up with a more accurate estimate of the Moon's position and velocity...the expected duration of transit is different for all three locations and depends on the Moon's exact location. So I might have to end up calculating the Moon's inclination too...

This graph shows the positions of the observers in inertial (Earth-centric, but non-rotating) space at each observation time. The markers are small, so you may have to zoom in, but triangles represent the California observation, circles the Minnesota one, and squares represent Derbyshire. The sphere represents the Earth, but its axial tilt is not represented visually. The X axis points away from the Sun and towards the Moon, which is assumed to be directly in the X axis at its previously measured distance at 0 UTC. The Y axis is the direction of motion of the Moon. The XY plane is the ecliptic plane, and Z is normal to the ecliptic. I think I may need to adjust the "time of day" angle of the Earth because California looks to be a little too close to sunset from this graph based on what actually happened, but the positions look correct relative to each other and the direction of Earth's axial tilt. Once I verify that the positions are accurate, I will be able to draw vectors between the observation positions and the observed direction of Mars through the Moon, and by finding where they cross I will locate Mars in 3D space.

I've been writing a MATLAB script to deal with these angles and rotations. Because the inertial positions of each observer are different at each observation, I need to calculate the new positions based on the location on Earth and the time elapsed. The motion of the observers is based on the rotation of Earth, which effectively changes the longitude of the observers as they move in a circle around the polar axis. The axis is at an angle to the ecliptic by 23.5 degrees, the axial tilt angle we're all familiar with. It is ALSO at an angle to the Sun-Earth-Moon line by a "season angle" which effectively quantifies where the Sun shines on the Earth and depends on the day of the year. To get the most accurate motion of the observers, both of these angles need to be accounted for to find the orientation of Earth at the times of observation. I could try to get a rough velocity vector for each observer and use that since it's not a huge amount of time so the motion of each observer will mostly be in the same direction, but I would need the position of the Moon in the sky from at least one location, and I didn't bother to gather that myself and I don't want to use data from an astronomy program except to check my work. Instead, I am using a sunrise/sunset table, which has been widely available for centuries...and MATLAB. I'll still write an explanation of all the calculations, it's all stuff that can be done by hand.

SpaceX launches what, half? Two-thirds? of all the orbital rockets of the world every year?

I've found something interesting. The first calculation I wanted to do was the angular velocity of Mars against the Moon. To do this, I measure the amount of time between Mars' disappearance and reappearance from my own location - simple enough. I decided to use the first pair of times in the sequence, since Mars' angular size would mess up the calculation otherwise. These would be the first and third times recorded. I got a difference of 3977 seconds. Then I decided to check my value against the other pair, the second and fourth times, expecting that they should match if the measurements are good. I got a value of 3968.8 seconds! That's 8.2 seconds shorter, which is a bit alarming for an error considering that's in the range of what I'm expecting to see for the parallax difference! I found that the amount of time it took Mars to rise was 8.2 seconds shorter than the time it took to set as well. The rise and set times should be the same length. I think there are a few things that could be going on here. The first option is that I just don't know what is happening to any precision smaller than 8.2 seconds, and I wasn't able to measure the timing well. Errors like this might throw off my calculation quite a bit. Because the time difference is the same amount in both measures I think that multiple data points are affected and not just one. Another possibility is that the Moon's non-roundness has struck again. Mars has some horizontal motion as it passes the Moon, which could cause surface variations to affect the exact timing of the occultation. A difference of a few seconds could occur if, say, Mars began to touch the Moon at the top of a mountain and then "slid" into a valley. Now, the 8.2 second difference is 0.2% of Mars' entire trip behind the Moon. This means that my estimate of Mars' angular velocity will be off by 0.2% - but Mars is very slow, so this may not completely wash out the parallax. There are a lot of seconds in the Atlantic Ocean, though, even at the Moon's blistering speed. We will see once I do the vector math on the observers' positions.

I'm an aerospace engineering student, and I can count all the women in my class on one hand... I'd like to meet more friends-of-friends, I think. And have more 'casual' interactions in general where I'm not just telling space facts the whole time. I never feel like I'm in an environment where it would be appropriate to flirt.