Space slingshot

From: "Richard Lamb"
Subject: Space slingshot was: Space Elevator
Date: Saturday, July 03, 2004 11:32 AM

Morgans wrote:


The thing that has me most concerned is cable aero drag. I think we
better
make that cable streamlined, and *real* thin, or drag could spell an end,
right off.
--
Jim in NC


Based on some very shakey assumptions...

Looking for drag per killometer of cable....

1) Using EAS for speed (already corrected for temperature and pressure
at cruise altitude)
218 K EAS = 250 MPH

2) that makes Q = 160 (.00256 * V^2 where V in MPH)

3) RN = 7657 Million (that's for 1 KM chord - in line with slipstream)
Viscosity effects predominate on the part of the cable in trail!
But I have no clue how to calculate the drag on that part...

4) For the part of the cable that is _perpendicular_ to the
slipstream...

Dp = Cd S Q
Cd = .02 ?? for a round cross section
Cable diameter guessed at 2 inches diameter
S = 546 sq feet per Km of cable length

Dp = 1750 pounds per Km of cable exposed perpendicular to the stream.


Well?

I'm more a concepts person, not math, g so if you are right, or even
close, that is not so bad. Next, we need to estimate the weight of the
cable, so we will know how much it is going to droop, (calculus anyone?),
and see how much of the weight is going to be felt at the orbiters, in both
vectors.

--
Jim in NC


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Morgans wrote:

From: "Richard Lamb"
Subject: Space slingshot was: Space Elevator
Date: Saturday, July 03, 2004 11:32 AM

Morgans wrote:


The thing that has me most concerned is cable aero drag. I think we
better
make that cable streamlined, and *real* thin, or drag could spell an end,
right off.
--
Jim in NC


Based on some very shakey assumptions...

Looking for drag per killometer of cable....

1) Using EAS for speed (already corrected for temperature and pressure
at cruise altitude)
218 K EAS = 250 MPH

2) that makes Q = 160 (.00256 * V^2 where V in MPH)

3) RN = 7657 Million (that's for 1 KM chord - in line with slipstream)
Viscosity effects predominate on the part of the cable in trail!
But I have no clue how to calculate the drag on that part...

4) For the part of the cable that is _perpendicular_ to the
slipstream...

Dp = Cd S Q
Cd = .02 ?? for a round cross section
Cable diameter guessed at 2 inches diameter
S = 546 sq feet per Km of cable length

Dp = 1750 pounds per Km of cable exposed perpendicular to the stream.

Well?

I'm more a concepts person, not math, g so if you are right, or even
close, that is not so bad. Next, we need to estimate the weight of the
cable, so we will know how much it is going to droop, (calculus anyone?),
and see how much of the weight is going to be felt at the orbiters, in both
vectors.

--
Jim in NC


Sorry Jim, I need to protect my amateur status in case aircraft design
is ever included in Olympic competition!

Richard