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Topic: Engine calculations  (Read 2481 times)

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Aenigma

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Engine calculations
« on: January 26, 2003, 08:43:43 am »
Because the sheet by Semper did not show the engine output, i took on the task to make nice calculations on them to get some data on them.
Engines are splitted in two groups : from 1 to 5 and from 6 to 10. The Klingon statistics have already been completed, and i'm working on the Romulans right now.

For Klingon engine range 1 to 5 i found the following formula:
R*(6.17818*10^-9*M^4-2.085*10^-5*M^3+0.02566537*M^2-13.164414*M+2741.939029)=output
R is the skill of the engineering officer (lvl1 =1.000, lvl3 =1.500, lvl4 =1.572, lvl5= 1.715, lvl6= 2.000)
M is the engine mass.

For Klingon engine range 6 to 10 the following formula:
R*(-3.91*10^-11*M^4+3.1637*10^-7*M^3-8.8532*10^-4*M^2+1.3972209*M-126.66137)=output

Aenigma  
« Last Edit: January 26, 2003, 09:12:26 am by Aenigma »
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Aenigma

  • Guest
Re: Engine calculations
« Reply #1 on: January 26, 2003, 08:52:22 am »
The data on the Romulan engine has been completed as well.
 EDIT: Not really i made a mistake on the mass for engine 1
For Romulan engine range 6 to 10
R*(1.0998*10^-10*M^4-9.858*10^-7*M^3+3.27068*10^-3*M^2-4.3305513M+2689.11779)=output

I'll be posting borg and federation in the coming week (don't know when i have time to do this).

Aenigma  
« Last Edit: January 26, 2003, 09:13:25 am by Aenigma »
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Aenigma

  • Guest
Re: Engine calculations
« Reply #2 on: January 26, 2003, 09:37:38 am »
New updated information.
Calculation for engines 1 to 5
For Romulan engine range 1 to 5
R*(1.7987*10^-8*M^4-4.5285*10^-5*M^3+0.0402709102*M^2-14.025658*M+1786.623)=output  
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Fornax

  • Guest
Re: Engine calculations
« Reply #3 on: January 26, 2003, 11:56:30 am »
Nice - so this is a regression?
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Vertigo

  • Guest
Re: Engine calculations
« Reply #4 on: January 26, 2003, 02:47:29 pm »
Quote:

Because the sheet by Semper did not show the engine output, i took on the task to make nice calculations on them to get some data on them.
Engines are splitted in two groups : from 1 to 5 and from 6 to 10. The Klingon statistics have already been completed, and i'm working on the Romulans right now.

For Klingon engine range 1 to 5 i found the following formula:
R*(6.17818*10^-9*M^4-2.085*10^-5*M^3+0.02566537*M^2-13.164414*M+2741.939029)=output
R is the skill of the engineering officer (lvl1 =1.000, lvl3 =1.500, lvl4 =1.572, lvl5= 1.715, lvl6= 2.000)
M is the engine mass.

For Klingon engine range 6 to 10 the following formula:
R*(-3.91*10^-11*M^4+3.1637*10^-7*M^3-8.8532*10^-4*M^2+1.3972209*M-126.66137)=output

Aenigma  



I'm confused here, not knowing what Semper's sheet is all about.  Can you explain the purpose of these formulas?  I assume it is not actual warp output, but some averaging process?  
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Aenigma

  • Guest
Engine calculations
« Reply #5 on: January 26, 2003, 08:43:43 am »
Because the sheet by Semper did not show the engine output, i took on the task to make nice calculations on them to get some data on them.
Engines are splitted in two groups : from 1 to 5 and from 6 to 10. The Klingon statistics have already been completed, and i'm working on the Romulans right now.

For Klingon engine range 1 to 5 i found the following formula:
R*(6.17818*10^-9*M^4-2.085*10^-5*M^3+0.02566537*M^2-13.164414*M+2741.939029)=output
R is the skill of the engineering officer (lvl1 =1.000, lvl3 =1.500, lvl4 =1.572, lvl5= 1.715, lvl6= 2.000)
M is the engine mass.

For Klingon engine range 6 to 10 the following formula:
R*(-3.91*10^-11*M^4+3.1637*10^-7*M^3-8.8532*10^-4*M^2+1.3972209*M-126.66137)=output

Aenigma  
« Last Edit: January 26, 2003, 09:12:26 am by Aenigma »
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Aenigma

  • Guest
Re: Engine calculations
« Reply #6 on: January 26, 2003, 08:52:22 am »
The data on the Romulan engine has been completed as well.
 EDIT: Not really i made a mistake on the mass for engine 1
For Romulan engine range 6 to 10
R*(1.0998*10^-10*M^4-9.858*10^-7*M^3+3.27068*10^-3*M^2-4.3305513M+2689.11779)=output

I'll be posting borg and federation in the coming week (don't know when i have time to do this).

Aenigma  
« Last Edit: January 26, 2003, 09:13:25 am by Aenigma »
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Aenigma

  • Guest
Re: Engine calculations
« Reply #7 on: January 26, 2003, 09:37:38 am »
New updated information.
Calculation for engines 1 to 5
For Romulan engine range 1 to 5
R*(1.7987*10^-8*M^4-4.5285*10^-5*M^3+0.0402709102*M^2-14.025658*M+1786.623)=output  
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Fornax

  • Guest
Re: Engine calculations
« Reply #8 on: January 26, 2003, 11:56:30 am »
Nice - so this is a regression?
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Vertigo

  • Guest
Re: Engine calculations
« Reply #9 on: January 26, 2003, 02:47:29 pm »
Quote:

Because the sheet by Semper did not show the engine output, i took on the task to make nice calculations on them to get some data on them.
Engines are splitted in two groups : from 1 to 5 and from 6 to 10. The Klingon statistics have already been completed, and i'm working on the Romulans right now.

For Klingon engine range 1 to 5 i found the following formula:
R*(6.17818*10^-9*M^4-2.085*10^-5*M^3+0.02566537*M^2-13.164414*M+2741.939029)=output
R is the skill of the engineering officer (lvl1 =1.000, lvl3 =1.500, lvl4 =1.572, lvl5= 1.715, lvl6= 2.000)
M is the engine mass.

For Klingon engine range 6 to 10 the following formula:
R*(-3.91*10^-11*M^4+3.1637*10^-7*M^3-8.8532*10^-4*M^2+1.3972209*M-126.66137)=output

Aenigma  



I'm confused here, not knowing what Semper's sheet is all about.  Can you explain the purpose of these formulas?  I assume it is not actual warp output, but some averaging process?  
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