What could possibly go wrong?

On Jan 15, 2:37*pm, Peter von Tresckow wrote:
Ralph Jones wrote:
http://www.networkworld.com/community/node/82160

Aero tow is too expensive. NASA should build a giant
Pwinch. 80,000' winch
run should be enough *to get it close to 40k :-)

Pete

Lets see....that's 1280 lbs of 1/4 in. Spectra. With cable droop, no
tall obstacle for on15.0 miles. (Bonneville Salt flats or borrow
fifteen miles of interstate?) And engineers, how many h.p. winch to
pull 245 tons? (Electric/Diesel locomotive engine?)

Still sounds more appealing than 747 wake turbulence.



Karen

On Wednesday, January 16, 2013 11:31:56 AM UTC-5, Karen wrote:
And engineers, how many h.p. winch to
pull 245 tons? (Electric/Diesel locomotive engine?)

Locomotive launch! Awesome.

Just for fun, lets work out the numbers.

With that much tension, probably not too much catenary arc (droop). Taking advantage of the low specific gravity of Spectra (it floats) conducting the launch over water should be considered with the ground roll on a near-shore runway and the winch on a ship 15 miles or so offshore. Think Dillingham Field in Hawaii.

Presumably the rope tension will need to be equal to the glider weight as is the case with glider winches so 1/4" Spectra isn't going to do it. The German DAeC winch guidelines require a minimum of 150% the strength of the heaviest glider to be launched so the rope will need a breaking strength of no less than 1,275,000 Lbs-F. Consulting the Samson Rope "Amsteel Blue" catalog you will need 3 5/8" diameter rope.
________________________________
And engineers, how many h.p. winch to pull 245 tons? (Electric/Diesel locomotive engine?)
________________________________
We know the force in Lbs but we need the giant glider best climb speed (Vy) so assume 120 knots or 202 feet per second. A simple formula with give HP but overestimate since the rope speed will reach 202 FPS only at the beginning of the rotation phase but it helps estimate the peak HP required. One HP = speed in FPS times force in Lbs divided by 550 so 245 tons is 490,000 pounds times 202 FPS gives 98,980,000 divided by 550 = 179,964 HP. For most of the launch 'merely'100,000 or so HP would be enough if the ship based scheme scheme was used. If the ship were sailing away from the glider at the start of the launch, its kinetic energy would supply any "excess" HP needed at the start of the launch.

So what engine could do that? The Wärtsilä-Sulzer RTA96-C could.
http://en.wikipedia.org/wiki/W%C3%A4...Sulzer_RTA96-C

I'll estimate the launch will take about 10 minutes so with a specific fuel consumption of .28 Lbs of per HP/Hr the launch would consume about 520 gallons of heavy bunker oil or about what a 747 uses taxiing to the runway.
_________________________________
Still sounds more appealing than 747 wake turbulence.
________________________________

Indeed!

So what could go wrong? You'd need a well thought out "launch abort" plan so the glider could always be landed back on the departure runway if the launch fails very similar to the Space Shuttle abort procedure. Most likely, it would not be necessary to jettison the payload in an emergency.

Bill D

On Wednesday, January 16, 2013 11:28:47 AM UTC-6, Bill D wrote:
Just for fun, lets work out the numbers. With that much tension, probably not too much catenary arc (droop). Taking advantage of the low specific gravity of Spectra (it floats) conducting the launch over water should be considered with the ground roll on a near-shore runway and the winch on a ship 15 miles or so offshore. Think Dillingham Field in Hawaii. Presumably the rope tension will need to be equal to the glider weight as is the case with glider winches so 1/4" Spectra isn't going to do it. The German DAeC winch guidelines require a minimum of 150% the strength of the heaviest glider to be launched so the rope will need a breaking strength of no less than 1,275,000 Lbs-F. Consulting the Samson Rope "Amsteel Blue" catalog you will need 3 5/8" diameter rope. ________________________________ And engineers, how many h.p. winch to pull 245 tons? (Electric/Diesel locomotive engine?) ________________________________ We know the force in Lbs but we need the giant glider best climb speed (Vy) so assume 120 knots or 202 feet per second. A simple formula with give HP but overestimate since the rope speed will reach 202 FPS only at the beginning of the rotation phase but it helps estimate the peak HP required. One HP = speed in FPS times force in Lbs divided by 550 so 245 tons is 490,000 pounds times 202 FPS gives 98,980,000 divided by 550 = 179,964 HP. For most of the launch 'merely'100,000 or so HP would be enough if the ship based scheme scheme was used. If the ship were sailing away from the glider at the start of the launch, its kinetic energy would supply any "excess" HP needed at the start of the launch. So what engine could do that? The Wärtsilä-Sulzer RTA96-C could. http://en.wikipedia.org/wiki/W%C3%A4...Sulzer_RTA96-C I'll estimate the launch will take about 10 minutes so with a specific fuel consumption of .28 Lbs of per HP/Hr the launch would consume about 520 gallons of heavy bunker oil or about what a 747 uses taxiing to the runway. _________________________________ Still sounds more appealing than 747 wake turbulence. ________________________________ Indeed! So what could go wrong? You'd need a well thought out "launch abort" plan so the glider could always be landed back on the departure runway if the launch fails very similar to the Space Shuttle abort procedure. Most likely, it would not be necessary to jettison the payload in an emergency. Bill D

i think the coolest way to deal with a failed winch launch in this scenario would be to fire the rocket, with the glider attached. this would quickly get you the altitude needed to be able to easily make it back to a runway.

On Wednesday, January 16, 2013 10:38:02 AM UTC-7, Tony wrote:
On Wednesday, January 16, 2013 11:28:47 AM UTC-6, Bill D wrote:

Just for fun, lets work out the numbers. With that much tension, probably not too much catenary arc (droop). Taking advantage of the low specific gravity of Spectra (it floats) conducting the launch over water should be considered with the ground roll on a near-shore runway and the winch on a ship 15 miles or so offshore. Think Dillingham Field in Hawaii. Presumably the rope tension will need to be equal to the glider weight as is the case with glider winches so 1/4" Spectra isn't going to do it. The German DAeC winch guidelines require a minimum of 150% the strength of the heaviest glider to be launched so the rope will need a breaking strength of no less than 1,275,000 Lbs-F. Consulting the Samson Rope "Amsteel Blue" catalog you will need 3 5/8" diameter rope. ________________________________ And engineers, how many h.p. winch to pull 245 tons? (Electric/Diesel locomotive engine?) ________________________________ We know the force in Lbs but we need the giant glider best climb speed (Vy) so assume 120 knots or 202 feet per second. A simple formula with give HP but overestimate since the rope speed will reach 202 FPS only at the beginning of the rotation phase but it helps estimate the peak HP required. One HP = speed in FPS times force in Lbs divided by 550 so 245 tons is 490,000 pounds times 202 FPS gives 98,980,000 divided by 550 = 179,964 HP. For most of the launch 'merely'100,000 or so HP would be enough if the ship based scheme scheme was used. If the ship were sailing away from the glider at the start of the launch, its kinetic energy would supply any "excess" HP needed at the start of the launch. So what engine could do that? The Wärtsilä-Sulzer RTA96-C could. http://en.wikipedia.org/wiki/W%C3%A4...Sulzer_RTA96-C I'll estimate the launch will take about 10 minutes so with a specific fuel consumption of .28 Lbs of per HP/Hr the launch would consume about 520 gallons of heavy bunker oil or about what a 747 uses taxiing to the runway. _________________________________ Still sounds more appealing than 747 wake turbulence. ________________________________ Indeed! So what could go wrong? You'd need a well thought out "launch abort" plan so the glider could always be landed back on the departure runway if the launch fails very similar to the Space Shuttle abort procedure. Most likely, it would not be necessary to jettison the payload in an emergency. Bill D



i think the coolest way to deal with a failed winch launch in this scenario would be to fire the rocket, with the glider attached. this would quickly get you the altitude needed to be able to easily make it back to a runway..

Plausible as a last resort.

Lets look at the acceleration and rotation phase.

1G acceleration is 19 knots per seconds per second so let's set Vr at 100 kts and Vy at 120 kts. The glider would lift its nose wheel in 5.25 seconds and lift off in about 6 seconds. Vy would be achieved in 6.3 seconds. Vr would be reached in about 130 meters or 430 feet, Vy in about 200m or 656 feet. Assuming you start with a 10,000' runway, that leaves about 9300' left to land and stop straight ahead from 120 knots. You might be able to do this up to a height of 1200 feet AGL given good glide path control devices.. Of course, the runway would have an over run braking surface at the departure end plus maybe, arresting wires. Above that, if the glider L/D is high enough, a circle-to-land maneuver should be possible.

Do I think this is actually possible? With the right glider design, absolutely. I hope the Dillingham guys will put a winch on a boat and try for an altitude record as a proof of concept.

On Wednesday, January 16, 2013 12:14:36 PM UTC-6, Bill D wrote:

"Plausible as a last resort. Lets look at the acceleration and rotation phase. 1G acceleration is 19 knots per seconds per second so let's set Vr at 100 kts and Vy at 120 kts. The glider would lift its nose wheel in 5.25 seconds and lift off in about 6 seconds. Vy would be achieved in 6.3 seconds. Vr would be reached in about 130 meters or 430 feet, Vy in about 200m or 656 feet. Assuming you start with a 10,000' runway, that leaves about 9300' left to land and stop straight ahead from 120 knots. You might be able to do this up to a height of 1200 feet AGL given good glide path control devices. Of course, the runway would have an over run braking surface at the departure end plus maybe, arresting wires. Above that, if the glider L/D is high enough, a circle-to-land maneuver should be possible. Do I think this is actually possible? With the right glider design, absolutely. I hope the Dillingham guys will put a winch on a boat and try for an altitude record as a proof of concept."

Minor and forgivable glitch, Bill. Knots is already velocity. Knots per second is acceleration. You said knots per second per second. That is rate of change of acceleration, or "onset". I won't use the other term, as some might think I am snarking at you.

Your boat will have to be anchored, or there will be many HP expended keeping it in place during the climb. So, no initial launch acceleration by having the boat moving. Dang it.

Ture airspeed as altitude increases? 200 feet per second true going through FL200 won't be much indicated airspeed. So, your rope speed will keep going up as the climb progresses. Try to take advantage of the wind aloft. Maybe you can reduce this effect?

You probably will start seeing significant horsepower lost due to the rope being dragged through the air by the glider, and it probably will have a lot more of a catenary. Yeah, tension will be high, but so will drag. And also, the weight will not be insignificant as far as power to lift the rope goes. How much will 15 miles of that 3.625 inch Amsteel blue weigh?

Things change a lot when looking at the horsepower required to lift and drag 20 lbs versus thousands of pounds. Percentages may be nearly the same, but absolute values are not. I think your fuel consumption estimate is low, Bill. :-)

And while we are talking of far out concepts, I still like the idea of being above the jet stream and dropping a chute on a long rope down into it. It starts out falling behind you, but once the chute is in the high velocity "core" of the jetstream, you turn around, allow the chute to inflate and if the gradient is high enough and your plane can fly slow enough, the chute can drag you "downwind."

On Wednesday, January 16, 2013 2:52:14 PM UTC-7, Steve Leonard wrote:
On Wednesday, January 16, 2013 12:14:36 PM UTC-6, Bill D wrote:



"Plausible as a last resort. Lets look at the acceleration and rotation phase. 1G acceleration is 19 knots per seconds per second so let's set Vr at 100 kts and Vy at 120 kts. The glider would lift its nose wheel in 5.25 seconds and lift off in about 6 seconds. Vy would be achieved in 6.3 seconds. Vr would be reached in about 130 meters or 430 feet, Vy in about 200m or 656 feet. Assuming you start with a 10,000' runway, that leaves about 9300' left to land and stop straight ahead from 120 knots. You might be able to do this up to a height of 1200 feet AGL given good glide path control devices. Of course, the runway would have an over run braking surface at the departure end plus maybe, arresting wires. Above that, if the glider L/D is high enough, a circle-to-land maneuver should be possible. Do I think this is actually possible? With the right glider design, absolutely. I hope the Dillingham guys will put a winch on a boat and try for an altitude record as a proof of concept."



Minor and forgivable glitch, Bill. Knots is already velocity. Knots per second is acceleration. You said knots per second per second. That is rate of change of acceleration, or "onset". I won't use the other term, as some might think I am snarking at you.



Your boat will have to be anchored, or there will be many HP expended keeping it in place during the climb. So, no initial launch acceleration by having the boat moving. Dang it.



Ture airspeed as altitude increases? 200 feet per second true going through FL200 won't be much indicated airspeed. So, your rope speed will keep going up as the climb progresses. Try to take advantage of the wind aloft. Maybe you can reduce this effect?



You probably will start seeing significant horsepower lost due to the rope being dragged through the air by the glider, and it probably will have a lot more of a catenary. Yeah, tension will be high, but so will drag. And also, the weight will not be insignificant as far as power to lift the rope goes. How much will 15 miles of that 3.625 inch Amsteel blue weigh?



Things change a lot when looking at the horsepower required to lift and drag 20 lbs versus thousands of pounds. Percentages may be nearly the same, but absolute values are not. I think your fuel consumption estimate is low, Bill. :-)



And while we are talking of far out concepts, I still like the idea of being above the jet stream and dropping a chute on a long rope down into it. It starts out falling behind you, but once the chute is in the high velocity "core" of the jetstream, you turn around, allow the chute to inflate and if the gradient is high enough and your plane can fly slow enough, the chute can drag you "downwind."

Thanks for the comments, Steve. Of course, you're right 1G is 19 knots/sec..

However, rope speed must sharply be reduced as the glider enters the climb - and even more if there are headwinds otherwise the glider over speeds. I suspect the true airspeed effect would be more than offset by upper winds..

15 miles = 269,280 lbs However, since the rope is getting shorter as you climb, you wouldn't have to lift that much. Drag is more important than weight. You'll gain a bit of Cd on Reynolds number compared to the thin .188" diameter rope commonly used in glider winches.

Actually, I assumed full HP for the whole launch but it will have to be backed way off in the upper half. Tension stays constant but HP doesn't since rope speed has to drop to avoid over speeding the glider. (Water skier effect) I think I probably over estimated the fuel.

A big ocean going tug like the "Far Sampson" could be the ideal platform. It could maintain position with its monster engines. If that turns out to be a problem, a big sea anchor would be the next option. The big Sultzer diesel might not be the best choice for 10 minutes of power. A gas turbine might be a better choice.

On Wednesday, January 16, 2013 6:06:59 PM UTC-6, Bill D wrote:
On Wednesday, January 16, 2013 2:52:14 PM UTC-7, Steve Leonard wrote:

On Wednesday, January 16, 2013 12:14:36 PM UTC-6, Bill D wrote:







"Plausible as a last resort. Lets look at the acceleration and rotation phase. 1G acceleration is 19 knots per seconds per second so let's set Vr at 100 kts and Vy at 120 kts. The glider would lift its nose wheel in 5.25 seconds and lift off in about 6 seconds. Vy would be achieved in 6.3 seconds. Vr would be reached in about 130 meters or 430 feet, Vy in about 200m or 656 feet. Assuming you start with a 10,000' runway, that leaves about 9300' left to land and stop straight ahead from 120 knots. You might be able to do this up to a height of 1200 feet AGL given good glide path control devices. Of course, the runway would have an over run braking surface at the departure end plus maybe, arresting wires. Above that, if the glider L/D is high enough, a circle-to-land maneuver should be possible. Do I think this is actually possible? With the right glider design, absolutely. I hope the Dillingham guys will put a winch on a boat and try for an altitude record as a proof of concept."







Minor and forgivable glitch, Bill. Knots is already velocity. Knots per second is acceleration. You said knots per second per second. That is rate of change of acceleration, or "onset". I won't use the other term, as some might think I am snarking at you.







Your boat will have to be anchored, or there will be many HP expended keeping it in place during the climb. So, no initial launch acceleration by having the boat moving. Dang it.







Ture airspeed as altitude increases? 200 feet per second true going through FL200 won't be much indicated airspeed. So, your rope speed will keep going up as the climb progresses. Try to take advantage of the wind aloft. Maybe you can reduce this effect?







You probably will start seeing significant horsepower lost due to the rope being dragged through the air by the glider, and it probably will have a lot more of a catenary. Yeah, tension will be high, but so will drag. And also, the weight will not be insignificant as far as power to lift the rope goes. How much will 15 miles of that 3.625 inch Amsteel blue weigh?







Things change a lot when looking at the horsepower required to lift and drag 20 lbs versus thousands of pounds. Percentages may be nearly the same, but absolute values are not. I think your fuel consumption estimate is low, Bill. :-)







And while we are talking of far out concepts, I still like the idea of being above the jet stream and dropping a chute on a long rope down into it.. It starts out falling behind you, but once the chute is in the high velocity "core" of the jetstream, you turn around, allow the chute to inflate and if the gradient is high enough and your plane can fly slow enough, the chute can drag you "downwind."



Thanks for the comments, Steve. Of course, you're right 1G is 19 knots/sec.



However, rope speed must sharply be reduced as the glider enters the climb - and even more if there are headwinds otherwise the glider over speeds. I suspect the true airspeed effect would be more than offset by upper winds.



15 miles = 269,280 lbs However, since the rope is getting shorter as you climb, you wouldn't have to lift that much. Drag is more important than weight. You'll gain a bit of Cd on Reynolds number compared to the thin ..188" diameter rope commonly used in glider winches.



Actually, I assumed full HP for the whole launch but it will have to be backed way off in the upper half. Tension stays constant but HP doesn't since rope speed has to drop to avoid over speeding the glider. (Water skier effect) I think I probably over estimated the fuel.



A big ocean going tug like the "Far Sampson" could be the ideal platform. It could maintain position with its monster engines. If that turns out to be a problem, a big sea anchor would be the next option. The big Sultzer diesel might not be the best choice for 10 minutes of power. A gas turbine might be a better choice.

Can't help myself but I have to add a 'political' comment: If we were on the metric system errors as made above would be unlikely. When oh when will this country ditch the silly medieval units of measure we use (and nobody else) and join the world and scientific community?
Herb

On Jan 16, 6:06*pm, Bill D wrote:

Thanks for the comments, Steve. *Of course, you're right 1G is 19 knots/sec.

However, rope speed must sharply be reduced as the glider enters the climb - *and even more if there are headwinds otherwise the glider over speeds. *I suspect the true airspeed effect would be more than offset by upper winds.

15 miles = 269,280 lbs *However, *since the rope is getting shorter as you climb, you wouldn't have to lift that much. *Drag is more important than weight. *You'll gain a bit of Cd on Reynolds number compared to the thin .188" diameter rope commonly used in glider winches.

Actually, I assumed full HP for the whole launch but it will have to be backed way off in the upper half. *Tension stays constant but HP doesn't since rope speed has to drop to avoid over speeding the glider. (Water skier effect) *I think I probably over estimated the fuel.

A big ocean going tug like the "Far Sampson" could be the ideal platform. It could maintain position with its monster engines. *If that turns out to be a problem, a big sea anchor would be the next option. *The big Sultzer diesel might not be the best choice for 10 minutes of power. *A gas turbine might be a better choice.


Rope speed must be reduced on a NORMAL winch launch, where there is
not much density change between start and ending altitude. Remember.
This is NOT a normal launch we are talking about. And the wind
doesn't always blow, so you won't always get that effective horsepower
and the winch speed may have to increase at some point in the launch.
Agree that speed will hit a high right at rotate, and slow some from
there. But when you are talking of climbing to where air density is
just a fraction of what it is at launch, you may well be reeling in
the rope considerably faster for the last half of the launch than you
are at lift-off.

There is also the matter of accelerating a quarter of a million pounds
of rope in addition to the half million pounds of plane. And by the
way, the article references a paylod of 490,000. That is not the
total weight of the glider and payload package! Ugh! In your normal
winch launch of a 1000 lb glider, you aren't having to accelerate 500
lbs of rope, so you have an easier time accelerating the system.
There are scaling effects that have been missed in the first brush.
But, that is OK. It is fun to think about.

And you may get a Cd advantage, but you are still dragging something
through the air that is almost 20 times the diameter at over twice the
glider flight speed. So, your Cd may be down a hair, but your total
drag is going to go up a bunch. Even in terms of percentage of total
system. Again with that scaling. And again, drag is likely more
important than weight when the weight of the rope is less than 1 or 2%
of the weight of the glider. When the weight of the rope is 50% of
the weight of the glider, it is a whole other story.

Still think you have under-estimated the fuel consumed.

Now, how can you possibly claim that if the boat is using its engines
to hold position, that it is not fuel consumed for the launch? Just
drop the anchor and be done with that part of it! :-)

Enough of this semi-technical talk. This is, after all,
Recreation.Aviation.Soaring. The place where people bash the SSA
Rules committe for not adopting the IGC Rules, talk about how ugly the
PW-5 is, and in general, just waste time.

Steve
(Kill-Joy)

Maintain high tow position...


"Karen" wrote in message
...
On Jan 15, 2:37 pm, Peter von Tresckow wrote:
Ralph Jones wrote:
http://www.networkworld.com/community/node/82160

Aero tow is too expensive. NASA should build a giant
Pwinch. 80,000' winch
run should be enough to get it close to 40k :-)

Pete

Lets see....that's 1280 lbs of 1/4 in. Spectra. With cable droop, no
tall obstacle for on15.0 miles. (Bonneville Salt flats or borrow
fifteen miles of interstate?) And engineers, how many h.p. winch to
pull 245 tons? (Electric/Diesel locomotive engine?)

Still sounds more appealing than 747 wake turbulence.



Karen