NASA formula, can someone decipher

[Don Shepherd]

See the attached formula from an IBM employee newsletter from 1969 related to the space program. Does anyone know what all this (the delta t) means?

[toml_12953]

(04-18-2020 08:42 PM)Don Shepherd Wrote:  See the attached formula from an IBM employee newsletter from 1969 related to the space program. Does anyone know what all this (the delta t) means?

Delta means change and t is temperature (in NASA-speak) so Δt would be the change in temperature. Wdot is Corrected Airflow per Area https://www.grc.nasa.gov/WWW/k-12/airplane/wcora.html

[Don Shepherd]

thanks Tom. Amazing stuff.

[Csaba Tizedes]

(04-18-2020 08:53 PM)toml_12953 Wrote:  

(04-18-2020 08:42 PM)Don Shepherd Wrote:  See the attached formula from an IBM employee newsletter from 1969 related to the space program. Does anyone know what all this (the delta t) means?

Delta means change and t is temperature (in NASA-speak) so Δt would be the change in temperature. Wdot is Corrected Airflow per Area https://www.grc.nasa.gov/WWW/k-12/airplane/wcora.html

Sound good, but I am sure, this is not correct.

The right side of equation inside the parentheses is look like a rise of an amount: (1-exp(-(...))) is a typical charging process equation - and dimensionless.
And the W.../W... is looks like a dimensionless factor, I guess the two W's have same dimension, therefore the right side is dimensionless.

So, it is looks like an amount how reached the final value - in dimensionless form or something time constant stuff, solution something first order ODE.

Csaba

[toml_12953]

(04-19-2020 08:52 PM)Csaba Tizedes Wrote:  

(04-18-2020 08:53 PM)toml_12953 Wrote:  Delta means change and t is temperature (in NASA-speak) so Δt would be the change in temperature. Wdot is Corrected Airflow per Area https://www.grc.nasa.gov/WWW/k-12/airplane/wcora.html

Sound good, but I am sure, this is not correct.

The right side of equation inside the parentheses is look like a rise of an amount: (1-exp(-(...))) is a typical charging process equation - and dimensionless.
And the W.../W... is looks like a dimensionless factor, I guess the two W's have same dimension, therefore the right side is dimensionless.

So, it is looks like an amount how reached the final value - in dimensionless form or something time constant stuff, solution something first order ODE.

Csaba

Did you look at the link? That's right from NASA. the t may either time or temperature but the definitions of delta and Wdot are exactly right.

[Don Shepherd]

Attached is the full page that I made this image from, maybe that might clarify things.

[twdeckard]

I'll throw in a few discordant guesses.

The e sin E - peaking thru is Keplers equation.

Wdot (first derivative of W, is the rate of change of vehicle or propellant weight)
Wb is the baseline weight of the vehicle at launch or at insertion.
G is gravity to turn the weights into masses.
The exponential decay is the propellant mass fraction
I say the lower case t is time

For a given change in velocity how much time elapses? a guess.

Todd

[Csaba Tizedes]

to Tom L: "Did you look at the link? That's right from NASA."
Sure, but I do not want to believe, there is someone at the NASA who is able to put an OPERATION (differentiation) into an INDEX... What's next? Using imperial units during the Mars-expedition?

to Tom L: "the t may either time or temperature"
Why do you think this is right after the above silly thing? I can show you something: \( \text{Cucumber}=\text{Fish} \cdot \text{Strawberry}^{2} \) this was the original equation from A. Einstein paper, but later modified to the well known \( \text{E}=\text{m} \cdot \text{c}^{2} \) .
Yeah, I know the "t" is typically temperature or time (because of greek or latin history, not English), but if you do not check the UNITS your equation maybe wrong.

to Tom L and twdeckard: "Wdot (first derivative of W, is the rate of change of vehicle or propellant weight) Wb is the baseline weight of the vehicle at launch or at insertion."

---UPDATE HERE
In this case the right side dimensions are \( \frac{\text{mass}}{\text{mass}/\text{time}} \) which is \( \text{time} \), so the left side dimension is correct.

Cs.

[Stevetuc]

(04-18-2020 08:42 PM)Don Shepherd Wrote:  See the attached formula from an IBM employee newsletter from 1969 related to the space program. Does anyone know what all this (the delta t) means?

Both the temperature ratio and the pressure ratio(delta) have the subscript t.
In the key, the t subscript = total conditions and o subscript = reference conditions

[SlideRule]

A "clean" image with readable subtext available at …IBM.COM…Apollo-Moon-Landing

BEST!
SlideRule

[BruceH]

(04-20-2020 12:41 AM)Don Shepherd Wrote:  Attached is the full page that I made this image from, maybe that might clarify things.

Cool! Hopscotch for bright kids...

[Don Shepherd]

(04-20-2020 08:59 PM)BruceH Wrote:  

(04-20-2020 12:41 AM)Don Shepherd Wrote:  Attached is the full page that I made this image from, maybe that might clarify things.

Cool! Hopscotch for bright kids...

That about says it all, Bruce, thanks. Now THAT I can understand!

When men first walked on the moon in 1969, the world was as one. Wouldn't it be wonderful if that happened again now (the world being as one again, that is), spurred on by how to defeat this pandemic?

Don

[John Keith]

(04-20-2020 10:21 PM)Don Shepherd Wrote:  When men first walked on the moon in 1969, the world was as one. Wouldn't it be wonderful if that happened again now (the world being as one again, that is), spurred on by how to defeat this pandemic?

It would indeed be wonderful, but in 1969 the world was not exactly "as one". The moon landing and the rest of the "space race" were part of the rivalry between the US and the USSR. Both nations had an overabundance of nuclear missiles pointed at each other and political intrigue occurring around the world.

The divisions between nations and ideologies today may not be as stark as during the Cold War but they are many and there is much instability in the world. I certainly hope that the pandemic will bring people together but sadly that doesn't seem to be happening at the moment.

[tom234]

Interesting Art Work did those NASA employees do that...or DARPA...well it would of through anyone...Hidden figures formulas...Who's spying on who?

Just a thought I would right partials and can't see what's missing.

[harryp]

I read Don Eyle's book about programming the computer for the Lunar Module. Been a while but the explanation might be in there.

[Pekis]

(04-20-2020 02:35 AM)twdeckard Wrote:  I'll throw in a few discordant guesses.

The e sin E - peaking thru is Keplers equation.

Wdot (first derivative of W, is the rate of change of vehicle or propellant weight)
Wb is the baseline weight of the vehicle at launch or at insertion.
G is gravity to turn the weights into masses.
The exponential decay is the propellant mass fraction
I say the lower case t is time

For a given change in velocity how much time elapses? a guess.

Todd

Hello,

It seems it's the case, according to http://www.braeunig.us/space/propuls.htm , derivating from the Tsiolkovsky's rocket equation:

For many spacecraft maneuvers it is necessary to calculate the duration of an engine burn required to achieve a specific change in Velocity:

t = m0 / q * (1 - exp( -DeltaV / C)

Isp = F / ( q * g0) (specific impulse of a rocket)
C = Isp * g0 [=> F / q] (effective exhaust gas Velocity)
F is thrust, q is the rate of mass flow, and g0 is standard gravity (9.80665 m/s2) [on Earth]

If q is an alias for WDOT
If g0 is an alias for g
If F is an alias for TH * g (?)
If m0 is an alias for WB

Then [Delta]t is indeed WB / WDOT * (1 - exp( -WDOT * DeltaV / (g * TH))


PROBLEM 1.4

A 5,000 kg spacecraft is in Earth orbit traveling at a velocity of 7,790 m/s.
Its engine is burned to accelerate it to a velocity of 12,000 m/s placing it
on an escape trajectory. The engine expels mass at a rate of 10 kg/s and an
effective velocity of 3,000 m/s. Calculate the duration of the burn.

SOLUTION,

Given: M = 5,000 kg
q = 10 kg/s
C = 3,000 m/s
DeltaV = 12,000 - 7,790 = 4,210 m/s

Equation (1.21),

t = M / q × [ 1 - 1 / e(DeltaV / C) ]
t = 5,000 / 10 × [ 1 - 1 / e(4,210 / 3,000) ]
t = 377 s

[jjohnson873]

Since we're on the subject of space, the moon and NASA, it is worth mentioning the NASA Artemis Program.