I have a question? e2e = 51.89 (Page 22).

[tom234]

Hi,
I was going through the D233 NASA paper.
I got the first part of page 22 first two answers...but the 3rd number still doesn't make sense to me.
I get that e2e=51.89 the closest I got was 56.421.
The great thing about solving on HP Prime is that equations have nice neat solutions...

Is there anyone that can find the correct answer as being e2e=51.89.
I will leave a video that was watched and the paper.
https://www.youtube.com/watch?v=78BdxTJ5spY&t=3s

If you get the time and maybe you can explain to me how they got that answer.

Thank You.

[Steve Simpkin]

That is a very interesting technical note from Nov 27, 1959. It has a number of examples of satellite de-orbit burn calculations. At the time, the average person would have had limited options to perform these calculations. A Slide Rule or tables come to mind. There is mention of using an IBM 704 computer. The IBM 704 was a large digital mainframe computer introduced by IBM in 1954. It was the first mass-produced computer with hardware for floating-point arithmetic. Changes from the 701 include the use of magnetic-core memory instead of Williams tubes.

In its day, the 704 was an exceptionally reliable machine. Being a vacuum-tube machine, however, the IBM 704 had very poor reliability by today's standards. On average, the machine failed around every 8 hours.

My favorite quote (quite an understatement).

“With a $2M price tag and weighing over 30,000 lbs, the IBM 704 was not a casual purchase.”

https://en.m.wikipedia.org/wiki/IBM_704

[EdS2]

Just a few notes, no helpful answer from me sorry...

θ angle in orbital plane measured from perigee point; also, true anomaly
subscript 2e equivalent selected position in first orbit

(where equivalent means accounting for the rotation of the earth since launch)

(we can even write the subscripts like so: θ₂ₑ)

So the worked example on p20 of the document (pdf page 22) is

Case A, eastward launch.- For case A, eastward launch, the given launch position is ϕ₁=28.50° N., λ₁=279.45° E.; the selected position is ϕ₂=34.00° N., λ₂=241.00° E.; the number of orbital passes n is 3; and the equivalent selected longitude λ₂ₑ is 309.689° E. The values of the first approximations are
[t(θ₂ₑ) - t(θ₁) ] ≈ 7.693
Δλ₁₋₂ₑ = 32.162
θ₂ₑ = 51.89
t(θ₂ₑ) = 12.702
ψ₁ = 70.468
i = 34.081

From Appendix B, page 18, we have
cos(θ₂ₑ - θ₁) = sinϕ₂ sinϕ₁ + cosϕ₂ cosϕ₁ cosΔλ₁₋₂ₑ

[tom234]

(02-23-2023 07:53 AM)EdS2 Wrote:  Just a few notes, no helpful answer from me sorry...

θ angle in orbital plane measured from perigee point; also, true anomaly
subscript 2e equivalent selected position in first orbit

(where equivalent means accounting for the rotation of the earth since launch)

(we can even write the subscripts like so: θ₂ₑ)

So the worked example on p20 of the document (pdf page 22) is

Case A, eastward launch.- For case A, eastward launch, the given launch position is ϕ₁=28.50° N., λ₁=279.45° E.; the selected position is ϕ₂=34.00° N., λ₂=241.00° E.; the number of orbital passes n is 3; and the equivalent selected longitude λ₂ₑ is 309.689° E. The values of the first approximations are
[t(θ₂ₑ) - t(θ₁) ] ≈ 7.693
Δλ₁₋₂ₑ = 32.162
θ₂ₑ = 51.89
t(θ₂ₑ) = 12.702
ψ₁ = 70.468
i = 34.081

From Appendix B, page 18, we have
cos(θ₂ₑ - θ₁) = sinϕ₂ sinϕ₁ + cosϕ₂ cosϕ₁ cosΔλ₁₋₂ₑ

I have done this but still can't get that number.

                   

[tom234]

(02-23-2023 12:11 PM)tom234 Wrote:  

(02-23-2023 07:53 AM)EdS2 Wrote:  Just a few notes, no helpful answer from me sorry...

θ angle in orbital plane measured from perigee point; also, true anomaly
subscript 2e equivalent selected position in first orbit

(where equivalent means accounting for the rotation of the earth since launch)

(we can even write the subscripts like so: θ₂ₑ)

So the worked example on p20 of the document (pdf page 22) is

Case A, eastward launch.- For case A, eastward launch, the given launch position is ϕ₁=28.50° N., λ₁=279.45° E.; the selected position is ϕ₂=34.00° N., λ₂=241.00° E.; the number of orbital passes n is 3; and the equivalent selected longitude λ₂ₑ is 309.689° E. The values of the first approximations are
[t(θ₂ₑ) - t(θ₁) ] ≈ 7.693
Δλ₁₋₂ₑ = 32.162
θ₂ₑ = 51.89
t(θ₂ₑ) = 12.702
ψ₁ = 70.468
i = 34.081

From Appendix B, page 18, we have
cos(θ₂ₑ - θ₁) = sinϕ₂ sinϕ₁ + cosϕ₂ cosϕ₁ cosΔλ₁₋₂ₑ

I have done this but still can't get that number.

       

[EdS2]

I don't know what's going on, but do you have separation of phi and theta? It looks like you have a ϕ₁ where you should have a θ₁.

That said, I'm not at all sure what the value of θ₁ should be. Possibly one could work backwards from the given solution to find out, and then it would all be clear.

[tom234]

(02-23-2023 01:38 PM)EdS2 Wrote:  I don't know what's going on, but do you have separation of phi and theta? It looks like you have a ϕ₁ where you should have a θ₁.

That said, I'm not at all sure what the value of θ₁ should be. Possibly one could work backwards from the given solution to find out, and then it would all be clear.

yes that is correct I could not find the θ, it was so small typing excuse that I used it as angle just from the character menu on the keyboard of Hp above VARS.

I was just using fast put variables to get the answer.
TY

I did work backwards using the solver and got those numbers. 56.??? that's what puzzling me on the first time around this should 51.???

[tom234]

(02-23-2023 04:44 AM)Steve Simpkin Wrote:  That is a very interesting technical note from Nov 27, 1959. It has a number of examples of satellite de-orbit burn calculations. At the time, the average person would have had limited options to perform these calculations. A Slide Rule or tables come to mind. There is mention of using an IBM 704 computer. The IBM 704 was a large digital mainframe computer introduced by IBM in 1954. It was the first mass-produced computer with hardware for floating-point arithmetic. Changes from the 701 include the use of magnetic-core memory instead of Williams tubes.

In its day, the 704 was an exceptionally reliable machine. Being a vacuum-tube machine, however, the IBM 704 had very poor reliability by today's standards. On average, the machine failed around every 8 hours.

My favorite quote (quite an understatement).

“With a $2M price tag and weighing over 30,000 lbs, the IBM 704 was not a casual purchase.”

https://en.m.wikipedia.org/wiki/IBM_704

IBM 380s we used this is decades before my time.

[Massimo Gnerucci]

(02-23-2023 07:21 PM)tom234 Wrote:  IBM 380s we used this is decades before my time.

I know of IBM 360/370/390/3090/43xx, or S/34/36/38 never saw a 380 though. ;)

[tom234]

(02-23-2023 07:32 PM)Massimo Gnerucci Wrote:  

(02-23-2023 07:21 PM)tom234 Wrote:  IBM 380s we used this is decades before my time.

I know of IBM 360/370/390/3090/43xx, or S/34/36/38 never saw a 380 though.

I was probably a 390 that was 1976...so long ago....ty.