Cycloid Payload Delivery Mechanism for Mars - Moon

tahanson43206 · 2024-02-19 09:08:23

The Cycloid Skyhook Concept goes back at least 50 years. I first saw it in the L5 News.

This topic is offered for members to report on updates to the original concept, due to improvements in knowledge of materials and computer controls, as well as other technologies that may be in play....

Update 2024/02/26... The current version of animation of the Cycloid system is available here:
Link to movie goes here:
https://i.imgur.com/VPFn7ip.mp4

The quote below comes from the 40 Tons topic ...

tahanson43206 wrote:

This post reflects a session with BARD that ended up as a Google Image lookup.
I'm not sure if I can post a link to the image I'd like to show, but here is the citation quote:
Carpentier, Justin & Benallegue, Mehdi & Laumond, Jean-Paul. (2017).
On the centre of mass motion in human walking.
International Journal of Automation and Computing.
14. 1-10. 10.1007/s11633-017-1088-5.
The center of mass (CoM) is a key descriptor in the understanding and the analysis of bipedal locomotion. Some approaches are based on the premise that humans minimize the CoM vertical displacement. Other approaches express walking dynamics through the inverted pendulum model. Such approaches are contradictory in that they lead to two conflicting patterns to express the CoM motion: straight line segments for the first approaches and arcs of a circle for the second ones. In this paper, we show that CoM motion is a trade-off between both patterns.
Specifically, CoM follows a "curtate cycloid", which is the curve described by a point rigidly attached to a wheel rolling on a flat surface. We demonstrate that all the three parameters defining a curtate cycloid only depend on the height of the subjects.
It is the cycloid I would like to show in this topic.
Link to image goes here:
Notes on image: The cycloid of interest for ** this ** topic is the one at the top in the image from the work by Carpentier, Justin & others.

(th)

tahanson43206 · 2024-02-19 09:09:03

This post is reserved for an index to posts that may be contributed by members over time.

(th)

tahanson43206 · 2024-02-19 09:14:25

This post is intended to provide a status report on study of the cycloid payload delivery concept in 2024.

In Sunday's Google Meeting, kdb512 reviewed the status of a Blender project to show the Cycloid Payload Delivery System in (simplified) operation.

He suggested adding a study of heating of the cable to the work plan, and I have done so.

The Blender model is in the final stages of preparation for animation. if all goes well, the movie will be translated to a GIF by imgur.com.

The animation shows the movement of a payload from orbit to ground delivery, following the Cycloid path as defined by a wheel "rolling" over the landscape.

Meanwhile, a Python program is in development, to compute (and show) forces at work on the cable between the space craft at the hub of the system, and the payload to be delivered.

As reported earlier, the heat load on the cable will be added to the Python development work plan.

(th)

tahanson43206 · 2024-02-19 19:32:29

This image is the last of 720 that make up a movie.

There are some steps between creating the 720 images and making a movie or GIF.

Here is a link to a movie that is stored on Dropbox:
https://www.dropbox.com/scl/fi/gdwtkun9 … d23ex&dl=0

After the file display, you can start the movie by clicking on the image, or clicking on the Play arrow at lower left.

And here, (if all goes well) is another view of the movie:
https://i.imgur.com/gV9O5MO.mp4

Please note that the imgur presentation shows the animation as a GIF (repeats) and it posts ads to the side. In the capitalist system those ads are what support the free service imgur provides.

What you are seeing is a (very simplified) animation of the Cycloid Payload Delivery system.

This system ** should ** work at Mars, but that is not yet clear. It ** definitely ** should work at the Moon, where atmosphere is not a factor.

I'm working on a Python program to compute the forces acting on the cable between the payload and the space craft at the hub of the wheel.

Please note that the actual connection between the space craft and the payload is a cable of some kind. The wheel is there to show the nature of the math at work in this delivery system.

There is a very good reason the Cycloid Payload Delivery concept has not been put into use after 50 years. The cable would be 100 kilometers long, and it would have to withstand many tons of tension in addition to the 40 tons of payload. Such a cable does not exist on Earth, and it may never exist anywhere.

The longest tension cable ever fabricated on Earth appears (according to Google) to be on the order of 3 kilometers long. A cable for the payload delivery application ** could ** be made by joining 100 sections of 1 kilometer each, so it isn't the length that is the problem. 20 tons is at the outer range of commercially supplied cables Google found for guy wires for massive radio towers. That problem could be solved by running cables in parallel. All this leads to the reasonable objection that the cable would be quite massive.

The classic solution to ** that ** problem is to balance the cable 100 kilometers on either side of the hub. In any case, this is likely to remain a theoretical exercise for another 50 years.

(th)

tahanson43206 · 2024-02-21 08:42:46

Circular orbit speed or velocity m/s
3500.10 m/s

The calculation above was performed by https://www.satsig.net/orbit-research/o … -speed.htm

The Cycloid Payload Delivery concept converts 3500.1 meters per second at 100 KM to 0 meters per second at 0 KM altitude.

It does this by rotating the payload along a path called a cycloid, as shown in posts above this one.

The delivery takes place in the time it takes the hub space craft to travel 100 kilometers along the orbit.

Per calculator, the time required is 28.5706122682 seconds.

The animation above is set to a time of 30 seconds.

(th)

GW Johnson · 2024-02-21 10:16:55

I'm finally getting a dim understanding of what this concept is about, after seeing the picture and the movie.

Things to consider:

1. 100 km is not high enough for an undecelerated craft to fly by Mars. The entry interface altitude for Mars is considered to be 135 km. Below that, at orbital-class speeds, the drag will claw you down.

2. You can balance the centrifugal force of the spinning cable "arm" with another one diametrically opposite. That is the same balancing mechanism as is claimed for the space elevator concept. The difference is that the space elevator does not spin (thus incurring no centrifugal forces, only varying gravity), it's just unbelievably enormous with a length about twice the synchronous orbit altitude.

3. The tip centrifugal acceleration is V^2/R = (3500 m/s)^2/(100,000 m) = 122.5 m/s^2, or some 12.49 standard gees. The cable itself feels half of that on average. The tension at the hub is going to be enormous, almost no matter how big the cable is, if for no other reason than the spin-accelerated payload force of 40 m.tons x 12.49 gees = 499.6 metric tons-force (or 4899 KN = 4.9 MN). The spin-accelerated mass of the cable adds to that and is just as extreme.

4. Unless the payload mass is trivial compared to the cable mass, you must release the same mass from the end of the balancing branch, just as you release from the payload delivery branch. Otherwise the vehicle at the center gets a sudden upward force equal to the payload mass times the tip acceleration, from the suddenly-unbalanced mass of the system. That force rotates around as the system rotates, too. That messes up its trajectory, and rather seriously.

5. As with the space elevator, there are no known and well-characterized materials from which to make the thing. I know the strength of a carbon nanotube is high, but no one has ever spun such fibers into a single thread, much less twisted or braided a cable out of thousands-to-millions of such threads. Such are held together by the friction of the spun (twisted-together) geometry that created the thread. What makes you think such friction will be higher just because the fibers are carbon nanotubes? I doubt VERY SERIOUSLY that if such a twine is ever actually made, that it will show a strength anywhere near the strength of the nanotubes it was made from! AND, it will be a lot harder to spin these nanotubes into a thread in the first place, because they are so much smaller than the fibers we normally spin into thread.

Just food for thought.

GW

tahanson43206 · 2024-02-21 10:37:40

For GW Johnson re #6

Thank you for taking a look at the animation!

Please keep this in mind for your projects. The animation you saw was about as minimalist as is possible, which is where I am right now. However, if you have something you'd like to animate for one of your presentations, I can presently handle up to three independently moving objects, and might be able to stretch to four, but that would be pushing things.

Regarding the several observations you made ... I'm right in the middle of something, and need to come back to study them.

The most valuable contribution I see in your post is the correction of the "safe" altitude for the space craft passing by Mars.

I got a velocity of 3500 meters per second for a satellite in orbit around Mars at 100 Km.

What would be the velocity of a space craft on a Hohmann orbit, with Mars coming up behind?

I assume the number is greater than 3500 M/s but am not sure. I ** know ** you've published the number several times in the forum and elsewhere, but we don't have a quick lookup capability, despite my efforts over the years to try create one.

Regarding 135 km vs 100 ... What drag / heating would the Hohmann space craft experience at 100 km? I assume / hope it is survivable.

I ask because the space craft has one duty and one duty only, to deliver the payload. It will take another long stretch to return to Earth, and if I recall your presentations correctly, Earth won't be there when the space craft arrives. However, that may not matter, if this is a commercial payload delivery service.

Regarding the balance on the cable .... I've thought about that, and appreciate your reminder.

My immediate next step is to work with ChatGPT4 to develop a Python program to compute the forces at work in the cable. I'm planning to provide options for the user of the program, to try various cable lengths, payload mass and other variables. kbd512 asked for temperature readings on the cable, and that is a reasonable suggestion, so I'll attempt to include it, but it would require fairly accurate figures for atmosphere density. Hopefully NASA and ESA have good data and hopefully it is available.

(th)

GW Johnson · 2024-02-21 14:19:13

Average Mars orbital speed is 24.121 km/s about the sun. At average planetary distances, the apohelion velocity on a Hohmann trajectory is about 21.499 km/s. The difference is the "Vfar" approximation for speed relative to Mars that is part of what sidesteps doing the 3-body problem: 2.622 km/s.

Mars gravity will accelerate the craft toward Mars as the two converge. Based on conservation of mechanical energy, the Vnear approximation is the square root of the sum of the squares of the Vfar value and Mars escape speed at the altitude of interest. This would be just barely less than the surface escape speed of 5.0282 km/s. Call it 4.955 km/s at 100 km altitude. This results in a "Vnear" of 5.606 km/s with respect to Mars at 100 km altitude.

The answer you asked for is 5.606 km/s wrt Mars at 100 km. That's essentially an atmospheric entry at somewhat over Mars escape speed at that altitude, and in fact bigger than surface escape speed. This estimate ignores the actual start of entry decel/heating, which begins at 135 km altitude. But it's in the ballpark, nonetheless.

GW

GW Johnson · 2024-02-21 14:25:08

Bear also in mind that the delivery craft is going to get accelerated around the curvature of Mars during all this. It will no longer be headed in the correct direction to stay on the Hohmann trajectory, even if it was undecelerated by the atmosphere of Mars. Directional changes are the very most expensive thing you can do in orbital mechanics. I have not run a detailed calculation on this problem, because that would require a 3-body finite element code. But an educated hunch says the delivery craft would need to do a delta vee on the order of that same 5.6 km/s in order to get back onto the original Hohmann trajectory. Earth will not be there when it arrives at perihelion 8.6 months later. But about some 3.5 years after launch, it will be. That's the synodic period estimate using Earth's period and the Hohmann orbit period.

GW

tahanson43206 · 2024-02-21 14:28:10

For GW Johnson!

Thank you for the value of velocity to be reckoned with on approach to Mars.

I just dropped off an estimate by Gemini.Google.com, of 700 meters per second as the Hohmann orbit component of the velocity.

That estimate does NOT take the pull of Mars on the space craft into account, as you have done. The figure of 5606 meters/second looks familiar, from the work you have published previously. Thank you!

It appears that an elevation of 135 kilometers and a velocity of 5600 m/s for the fly by with drop off scenario is what the Cycloid Delivery planner has to work with.

I'll make sure the Python program to compute Cycloid flight loads is able to cover those values.

(th)

tahanson43206 · 2024-02-21 14:33:59

For GW Johnson re #9

Thanks for mentioning the acceleration that will be imparted to a space craft that just flies by on a Hohmann trajectory! I was planning to ask about that, so I appreciate your including it in your post.

It seems to me that a mission planner with the appropriate software could find a path for the space craft that does something useful with the acceleration provided by Mars as it roars by the pokey space craft. Mission planners (are reported to) use such planetary encounters to give their space craft impulses of momentum that add up to useful inputs to a mission. In this case, we would like Mars to fling the space craft onto a path that brings it back to Earth sooner rather than later.

This topic is the Cycloid Payload Delivery topic, so I'll move this question over to Orbital Mechanics.

(th)

tahanson43206 · 2024-02-22 10:35:15

Details for performing computations for the test Cycloid include:

Test altitude: 100 km
Mars orbital velocity at 100 km: 3500 m/s
Radius of test wheel: 100 km
Diameter of test wheel: 200 km
Rotation rate of test wheel: 1/2 RPM (120 seconds per rotation)
Velocity of the payload at the tip of a spoke: 5.236 km/s

The payload must be accelerated from rest (ahead of the Hub space craft) to 3500 m/s at the surface, with vertical velocity of zero.

The time required to accelerate the payload is 30 seconds (as shown by the animation).

Near the conclusion of the rotation, the payload must be given an upward impulse by rockets at the hub.

At the successful finish of both accelerations, the payload will be released with zero horizontal and zero vertical velocity with respect to Mars.

The amounts of acceleration required is currently unknown.

The mass to be delivered is given as 40 tons.

(th)

tahanson43206 · 2024-02-22 14:06:49

Here is another look at the details for planning calculations using the test case.

In this presentation, ChatGPT4 has included an estimate of the amount of thrust that would be needed to rotate a 40 ton payload from ahead of the hub space craft to directly below the hub space craft in 28.57 seconds.

The animation showing the Cycloid Payload delivery process was set for 30 seconds, which is close enough for the purpose.

It should be noted that the payload is going to have to punch through 100 kilometers of Mars' atmosphere in 28.57 seconds, and the cable will make the same trip, depending upon the distance from the hub.

Summary of Rocket System Design for Circular Motion Exercise
We have explored a unique scenario involving the design of a rocket system to accelerate a mass along the circumference of a circle. Here is a concise summary of our findings and calculations:
- Objective: Design a rocket system to move a 40 metric ton mass from (+X=100 km, +Y=0) to (+X=0 km, -Y=100 km) along a circle with a radius of 100 kilometers. The movement completes a quarter turn in exactly 28.57 seconds.
- Initial Conditions: Lateral movement in the +X direction at 3500 m/s.
- Rotation Requirement: A quarter turn rotation where the point on the circumference initially at +X=100 and +Y=0 moves to +X=0 and -Y=100.
Key Calculations:
The time for the center of the wheel to move 100 kilometers at 3500 m/s was confirmed to be 28.57 seconds.
The rate of rotation required was calculated to be approximately 0.525 RPM.
The continuous thrust required to achieve this movement, assuming a mass of 40 metric tons, was calculated to be approximately 6,930,000 Newtons (N), with an acceleration of 173.25 m/s^2.
This exercise has provided valuable insights into the dynamics and requirements of designing a rocket system for a specific trajectory and rotation. It serves as a foundation for further exploration and refinement by rocket engineers and enthusiasts within our community.
Feel free to discuss, explore further, or raise questions based on this scenario and calculation.
Your contributions and insights are highly encouraged!

(th)

tahanson43206 · 2024-02-23 09:21:26

Feedback about the Cycloid animation arrived today from our Alaska correspondent, who's been out on a survey of residents of a very large geographic area.

Our correspondent observed that the animation takes 30 seconds, which seems slow (for an animation).

In fact, that is the ** ACTUAL ** speed of the flight, which means that it is extremely fast if you are a rocket engineer planning the flight.

The payload must be given continuous acceleration downward and then gradually more and more to the rear, because the payload must be given a 3500 meter per second impulse toward the rear, while transitioning to the surface for a vertical shift of 100 kilometers.

The 30 seconds is actually 28.57 seconds, which is the time required for the space craft at the hub to traverse 100 km, at 3500 meters per second.

While the payload is under continuous thrust, the hub itself is providing thrust to counter the forces at work as the payload moves down in accordance with the Cycloid curve. It is the amount of thrust needed at the hub that requires program calculation.

It will vary from zero at the beginning of the payload movement, to maximum at the instant before release of the payload at the surface.

(th)

tahanson43206 · 2024-02-26 18:18:54

This post is reserved for a link to the latest version of the Cycloid Payload Delivery Animation.

This version shows rocket exhaust at the payload, and at the hub, and the first draft of text reporting is defined.

Link to image goes here:

Above is the original image. See movie for enhancements.

Link to movie goes here:
https://i.imgur.com/VPFn7ip.mp4

(th)

tahanson43206 · 2024-03-02 08:08:19

Work on the Cycloid Payload Delivery concept continues ...

This post is about the rate of movement of the payload from orbit to the ground drop off point....

The animation shows a timeline of 30 seconds, which corresponds to the time that would be required if a pure cycloid curve were implemented. However, there is no need to implement a pure cycloid curve. That curve is useful for starting the thought process.

If the time is stretched, and if the atmosphere is enlisted to take the load off the cable, then the descent from orbit to the surface will take longer, but a gentle drop off will be achieved, and the Hub can collect the unloaded cable as it swings back up to orbit.

Here (again) is the current animation....
Update 2024/02/26... The current version of animation of the Cycloid system is available here:
Link to movie goes here:
https://i.imgur.com/VPFn7ip.mp4

Work currently underway is intended to display status messages during the run.

Update later: Once the concept of wings on the payload delivery pod is added to the vision, a logical extension is to add wings a other locations on the cable. These can assist in decreasing the load on the cable as well. The tradeoff is that they increase mass of the cable, and ultimately that means adjustment in the thrust needed at the Hub to restore equilibrium after a delivery.

(th)

New Mars Forums

Announcement

#1 2024-02-19 09:08:23