Numbers in the appendix of Arthur C. Clarke's 1945 classic paper

[MB2012]

Dear All,

This question relates to the numbers in the appendix of Arthur C. Clarke's 1945 classic paper (Wireless World, Oct. 1945, pp. 305-308).

The appendix (on page 308 of the paper) outlines some basic principles of rocket design, including Tsiolkovsky's fundamental equation of rocket motion which is given as:

V = v log_e (R)

where V is the final velocity of the rocket, v is the exhaust velocity, and R is the ratio of initial mass to final mass (payload plus structure).

Clarke states that: "If we assume v to be 3.3 km/sec. R will be 20 to 1." (For V = 10 km/sec.)

Indeed it will:

V = v * log_e (R)

10 = 3.3 * log_e (R)

R = e^(10/3.3)

R = 20.7

QED

So far, so good.

Clarke then states that due to the rocket's finite acceleration, it loses velocity as a result of gravitational retardation and the necessary ratio R_g is increased to

R_g = R * ( (a + g) / (a) )

where a is the rocket's acceleration and g is the acceleration due to gravity.

Clarke goes on to say that (and this is where my problem is): "For an automatically controlled rocket a would be about 5g and so the necessary R would be 37 to 1."

So, by my calculations:

R_g = R * ( (a + g) / (a) )

R_g = R * ( (5g + g) / (5g) )

R_g = R * (6g / 5g)

R_g = 1.2 * R

(EQUIVALENTLY: R = 0.83 * R_g)

How on earth (or perhaps more appropriately, how in space) does the necessary R come to be 37 to 1?

Where am I going wrong?

Thanks in advance.

MB

 

[RGClark]

I can't see where he gets that answer either. Would it be allowable subject to copywrite restrictions to scan that page and post it on the forum?

Bob Clark

 

[Hlynkacg]

Dear All,

This question relates to the numbers in the appendix of Arthur C. Clarke's 1945 classic paper (Wireless World, Oct. 1945, pp. 305-308).

The appendix (on page 308 of the paper) outlines some basic principles of rocket design, including Tsiolkovsky's fundamental equation of rocket motion which is given as:

V = v log_e (R)

where V is the final velocity of the rocket, v is the exhaust velocity, and R is the ratio of initial mass to final mass (payload plus structure).

Clarke states that: "If we assume v to be 3.3 km/sec. R will be 20 to 1." (For V = 10 km/sec.)

Indeed it will:

V = v * log_e (R)

10 = 3.3 * log_e (R)

R = e^(10/3.3)

R = 20.7

QED

So far, so good.

Clarke then states that due to the rocket's finite acceleration, it loses velocity as a result of gravitational retardation and the necessary ratio R_g is increased to

R_g = R * ( (a + g) / (a) )

where a is the rocket's acceleration and g is the acceleration due to gravity.

Clarke goes on to say that (and this is where my problem is): "For an automatically controlled rocket a would be about 5g and so the necessary R would be 37 to 1."

So, by my calculations:

R_g = R * ( (a + g) / (a) )

R_g = R * ( (5g + g) / (5g) )

R_g = R * (6g / 5g)

R_g = 1.2 * R

(EQUIVALENTLY: R = 0.83 * R_g)

How on earth (or perhaps more appropriately, how in space) does the necessary R come to be 37 to 1?

Where am I going wrong?

Thanks in advance.

MB

Given two values you can derive the third, I would assume (as I haven't read the paper) that Clark gave a target mass or dV somewhere in the paper given an acceleration of 5gs would require a MR of 37 to attain.

 

[MB2012]

Thank you for your responses.

Clarke's paper (“Extra-Terrestrial Relays: Can Rocket Stations Give World-Wide Radio Coverage?,” Wireless World, Oct. 1945, pp. 305–308) is reproduced in full on pages 16-22 of the following NASA e-book: http://history.nasa.gov/SP-4407/vol3/cover.pdf

The complete reference is: Logsdon, John M., ed., with Roger D. Launius, David H. Onkst, and Stephen J. Garber. Exploring the Unknown: Selected Documents in the History of the U.S. Civil Space Program, Volume III, Using Space. NASA SP-4407,1998.

I look forward to your further responses.

Many thanks,

MB

 

[Ares]

Hi, thanks for sharing this essay!

I think your problem is R_g = R to the power of, not multiplied by, ((a+g)/a). So R_g = 20 ^ 1.2 = 36.

 

[MB2012]

Hi, thanks for sharing this essay!

I think your problem is R_g = R to the power of, not multiplied by, ((a+g)/a). So R_g = 20 ^ 1.2 = 36.

Thank Ares.

That's interesting. I didn't consider that possibility (i.e. that it was an exponent, not a product).

I wonder whether this is just a typographical error in the original paper? Because the original paper does appear to show a product and not an exponent. Or whether there's some other explanation?

Incidentally, Clarke references the following book by Willy Ley for the formulae in his rocket design appendix: "Rockets: The future of travel beyond the stratosphere", 1944, New York, The Viking Press. I couldn't find a copy of the book to double-check and cross-reference the relevant formulae.

Can you or anyone confirm that the right/standard correction factor is indeed R_g = R ^ ( (a + g) / (a) ) and NOT R_g = R * ( (a + g) / (a) )? And could you please provide a technical reference for this?

Thanks again.

MB
----
a

 

[Ares]

Take a look at: [ame="http://en.wikipedia.org/wiki/Gravity_drag"]http://en.wikipedia.org/wiki/Gravity_drag[/ame] and http://www.projectrho.com/public_html/rocket/mission.php under the heading 'Drag'.

On the Atomic Rockets link, there is the equation: delta_vd = delta_esc * g / a, where delta_vd is the additional dv required to compensate for gravitational drag and delta_esc is escape velocity (or the total delta v without gravitational drag: 10 km/s in this case). I'm not sure where this comes from, but Atomic Rockets is a brilliant website so I'm inclined to trust it. I'd be interested if you can work it out though! Maybe google gravitational drag for a technical reference?

Then the rocket equation is: R = exp (V/v)
V = delta_v + delta_vd = 10 km/s + delta_esc * g / a
= 10 * (1+g/a) km/s
R = exp ( (10 kms/s * (1+g/a)) / v)
= exp (10 km/s / v) ^ (1+g/a)
= R0 ^ ((a+g)/a)
Where R is equivalent to Clarke's R_g and R0 is Clarke's R.

Does that make sense, or am I talking nonsense?

 

[MB2012]

Take a look at: http://en.wikipedia.org/wiki/Gravity_drag and http://www.projectrho.com/public_html/rocket/mission.php under the heading 'Drag'.

On the Atomic Rockets link, there is the equation: delta_vd = delta_esc * g / a, where delta_vd is the additional dv required to compensate for gravitational drag and delta_esc is escape velocity (or the total delta v without gravitational drag: 10 km/s in this case). I'm not sure where this comes from, but Atomic Rockets is a brilliant website so I'm inclined to trust it. I'd be interested if you can work it out though! Maybe google gravitational drag for a technical reference?

Then the rocket equation is: R = exp (V/v)
V = delta_v + delta_vd = 10 km/s + delta_esc * g / a
= 10 * (1+g/a) km/s
R = exp ( (10 kms/s * (1+g/a)) / v)
= exp (10 km/s / v) ^ (1+g/a)
= R0 ^ ((a+g)/a)
Where R is equivalent to Clarke's R_g and R0 is Clarke's R.

Does that make sense, or am I talking nonsense?

Thanks Ares, your contribution has really helped. It makes perfect sense.

Just a couple of fine points to round off the "proof":

* The "target" final velocity (V) in Clarke's appendix is the orbital velocity, NOT the escape velocity. (The escape velocity is 1.414 (root 2) times the orbital velocity: see Appendix 1 of Martin J. L. Turner, "Rocket and Spacecraft Propulsion", 2005, 2e, Springer.)

* The modified rocket equation, taking into account gravitational retardation, is (using Clarke's nomenclature): V = v log_e (R_g) - gt. (See Section 5.2 of Turner 2e)

* The second term "gt" is our delta_vd (i.e. additional dv required to compensate for gravitational drag), since a = dv/dt. So:

delta_vd = g * t = g * (V/a) = V * (g/a)

* Since V = v log_e (R_g) - gt and R = exp (V/v):

R_g = exp ((V + gt) / v)
= exp ((V + (V * (g/a))) / v)
= exp ((V * (1 + g/a)) / v)
= exp (V / v) ^ (1+g/a)
= R ^ ((a+g)/a)

* For a = 5g and R = 20:

R_g = R ^ ((a+g)/a)
= 20 ^ 1.2
= 37 (rounded up)
QED

 

[Ares]

Thanks for the PM! Does that assume that the rocket is travelling vertically then?

 

[Urwumpe]

Thanks for the PM! Does that assume that the rocket is travelling vertically then?

In that context: Yes (the gravity losses are be simply approximately g * t, and the orbital velocity useless: You have a periapsis radius of zero in any case).

 

[MB2012]

Thanks for the PM! Does that assume that the rocket is travelling vertically then?

That's correct, that particular form of the equation assumes vertical motion. But the simplified model serves its purpose: it provides insight.

Thanks again, one and all! You helped me to finally get rid of the pesky mindworm (a la earworm) that I got from Clarke's appendix.