Continuing the plotting and scheming. There have been no more flights, but…

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More numbers...

And they all mean something. I have been having some fun with this, every now and then, and it has been a source of some great relaxation and distraction during some pressured times, lately.

In the last post, I had got as far as calculating the mean orbital velocities based on orbital radius. This is just scratching the surface. But before I start, there was one thing I did on that last Mercury flight which I omitted (because the post was getting long inside the summary).

I had timed the orbit at five points: Northernmost point (which appeared to be abeam Cape Hatteras, giving me the approximate inclination of the orbit at about 33.5º), crossing the equator descending, the southern most point, crossing the equator ascending, and finally again at the northernmost point. This gave me the interval between those points, which helped me to approximate the eccentricity of the orbit. Straight to the table I compiled for that data…

As can be seen, I got the intervals of 21.8, 21.6, 22.7 and 22.3 minutes. The longer the interval, as covered in the last post, the longer the radius of the orbit in that segment. I multiplied each of these by 4 (as they represent a quarter orbit) to obtain a period value for a total, circular orbit, then converted it to seconds for calculation purposes. I used the formula…

*Radius = Cube Root of ( (Earth Mass kg * Grav Constant * Period Secs) / (4 * Pi^2) )*

To obtain the approximate radius at each segment. I had to “dumb this down” for Excel, after looking up why it does not do cube roots. This did the trick, which is the same thing…

*( (Earth Mass kg * Grav Constant * Period Secs) / (4 * Pi^2) ) ^(1 / 3)*

Anyway. I was able to determine to a degree the eccentricity of the orbit from this information, and in which segments were the apogee and perigee. And as before, I deduced the velocity from the circumference and period. I did not need much more from this table, for now.

On to the next thing. The Mercury capsule and its retro rockets. These are three, which fire for ten seconds each, in overlapping sequence, and each produce 4.5 Kilo-Newtons. Now, I would be wrong to use the value of 4.5 kN x 3, because the nozzles are pointed outwards, and it is a component of 4.5 kN that is directed along the longitudinal axis. Of course, they are angled out like that for a reason; they point at the CoG of the capsule, so that their firing does not induce a rotation. I made an estimate that they point outwards about 24.2º. That is;

*Arc Tangent ( 0.45 m / 1.0 m )*

That angle gave me a longitudinal thrust component of…

_Cos ( 24.2º ) x 4.5 kN = 4.1 kN

…times three, for a total thrust vector of 12.3 Kilo-Newtons for the deorbit. But there is more. The retro-burn of the Mercury is done with the back end of the capsule pointed outside of the the orbit path, planar, at 34º (apparently). This further divides those 12.3 kN into a retro burn component, and a radial burn component, which tips the orbit. In effect, for this experiment, it is enough to say that it induces some “earthward” vertical velocity. As it is never going to complete the orbit, this assumption is fair enough. By how much is here…

*Longitudinal component = Cos(34º) x 12.3 kN = 10.19 kN*

*Vertical component = Sin(34º) x 12.3 kN = 6.87 kN*

By how much does all that affect the capsule? We need the capsule’s momentum, for that. The capsule’s mass is 1,360 kg, and in the first segment, its velocity is 7,823.73 m/s. Multiply to get the longitudinal path momentum…

1,360 x 7,823.73 = 10,640,272.8 N. (1.06E7)

…subtract the longitudinal retro burn force times the 10 seconds (and times 1,000, don’t forget, kN) from it, to get the new capsule momentum…

*10,640,272.8 N - (10.19 N x 10 x 1,000) = 10,538,372.8 N*

Divide that by the capsule mass to get the new velocity…

*10,538,372.8 N / 1,360 kg = 7,748.8 m/s*

For the radial burn component, things are easier. As the craft is moving along the path, we can assume zero momentum relatively, and the force acts only on the mass to change its vertical velocity…

*(6,870 N x 10 secs) / 1,360 kg = 50.5 m/s.*

And that is the vertical (or “perpendicular to path”) velocity induced by the burn. This was important, by the way. Scott Carpenter’s deorbit was a bit off angle, and resulted in a considerably longer deorbit and reentry path.

Well. Here is the table…

There are a lot more numbers already generated on other tables, but this is all I have had time for tonight. It strikes me, while I have been playing with this, that the whole object, eventually, is to identify just one particular second during the orbits when to fire the retro rockets. That must have been the world’s most expensive second, in terms of effort invested. LOL!

All interesting stuff!

And a delightful little trick I stumbled upon while looking for the Excel cube root solution…