The NASA News item NASA’s New Shape-Shifting Radiator Inspired by Origami describes a technique for spacecraft temperature regulation combining both a shape-changing surface and an emissivity-changing material applied to it.

I don't understand what is really happening here. How can one shape be better for cooling (radiating into space) while another be better for warming (presumably from sunlight)? Wouldn't you just want the maximum area? It seems the deeper the folds, the smaller the total area exposed to the environment.

To reiterate, the deeper the folds, the smaller the cross-sectional area because the total amount of surface is fixed. Is it really true that a smaller cross-section with deeper folds is better than spread out and planar?

This novel radiator controls the rate of heat loss by performing shape-shifting maneuvers. The resulting topographical changes could be achieved with temperature-sensitive materials like muscle wire or shape-memory alloys. As temperature-sensitive materials experience a change in temperature — caused by spacecraft electronics or the absorption of heat from the Earth or sun — the radiator could automatically change its shape to either shed or conserve heat.

The deeper the folds or cavities, the greater the absorption, explained Mulford, adding that scientists have investigated the use of cavities to affect heat loss for nearly 100 years, but no one has approached the challenge in quite this way. “Origami allows you to change the depth of these cavities in real time, thereby changing the heat loss from a surface in real time,” he said.

below: "Brigham Young assistant professor Brian Iverson and doctoral student Rydge Mulford have teamed with NASA technologist Vivek Dwivedi to advance the design of a three-dimensional, foldable radiator, inspired by the art of paper folding. Still early in its development, Iverson and Mulford are experimenting with different shapes to determine which configuration would work best as a radiator." Credits: Brigham Young University enter image description here


I just saw this and recognized my research, ha ha. I realize this is an older question but I wanted to give my two cents. The published article is a bit deceitful in describing the technology (which frustrates me) so I wanted to straighten things out.

You are exactly right in that the finite surface area will offset almost any gains from increases in radiative surface properties. The first plot below shows how the apparent absorptivity and apparent emissivity of the surface openings will change as the V collapses from fully open (phi of 180 degrees) to fully closed (0 degrees) for a diffusely-reflecting surface. Each line is for a different intrinsic absorptivity that is inherent to the material. As you can see, the apparent absorptivity approaches one for all cases. However, as you have pointed out, the surface area of the radiator is decreasing. The net effect is that the total net radiative heat does indeed decrease as the surface collapses (see Figure 2 which is for the net radiative heat transfer of a diffusely-reflecting surface and several different intrinsic absorptivities). This still gives us significant control, though, over the heat transfer from the radiator just in the opposite direction that we originally intended. We are also exploring a variety of surfaces that either don't collapse to zero surface area or that maintain a given surface area while changing the cavity angle. See the presentation linked below for more of that information. I have also included Figure 3, which shows how the heat transfer of a specularly-reflecting (or mirror-like reflection) radiator behaves as the radiator collapses if the incident irradiation is collimated, or all coming from one direction (as is the case generally for solar irradiation). Here the multiple reflections combined with the singular direction of irradiation gives very erratic behavior and significant changes in heat transfer are possible over small actuation distances.

See slides 23 - 29 in this presentation for more of the math, experimental procedures, etc. Should have a paper soon talking about the behavior of net radiative heat transfer with actuation.

enter image description here Figure 1

enter image description here Figure 2

enter image description here Figure 3

  • $\begingroup$ Wow, that's great that you've stopped by and revived an old question! I'll give this and your link a read later today. It's always great to have people who "work in space" (so to speak) to provide input. I hope you continue to stop by and check out the other questions here as well! The thermal and thermal-control tags alone link to a few dozen questions. This one for example still needs a better answer posted: What are these very large, square panels on Inmarsat 5's? $\endgroup$ – uhoh Jan 25 '18 at 22:17

going on intuition: when a photon hits a surface, it's either absorbed or reflected. On a flat surface, this is easy: if the photon is reflected, it'll travel away from the spacecraft.
When a photon travels into a cavity, it'll have to be reflected multiple times to get out of the cavity again, so there are multiple chances for the photon to be absorbed. The absorption coefficient is multiplied by the average number of reflections needed to get out of the cavity.

  • $\begingroup$ I was thinking also, that the path of an object tends to be straight. In Folded Space as your saying any deflection/refraction etc. would not have a straight line out. $\endgroup$ – Enigma Maitreya May 6 '17 at 13:22
  • $\begingroup$ @Hobbes If the absorption coefficient is already high, say 0.8, and only half of the incident angles can geometrically result in a second incidence, then that's a 10% effect (0.8 + 0.16/2, actually its 1 - 0.2 - (0.2^2)/2). So OK it can be a small improvement for dark (high emissivity) things. It would be more helpful for low emissivity (light colored) things, but one wouldn't use those. Perhaps I'm misunderstanding the geometrical aspect - I thought it extends out flat or folds up such that folding would dramatically reduce total cross-section or Etendu (m^2 sR)? $\endgroup$ – uhoh May 6 '17 at 14:57
  • $\begingroup$ I have a hunch that this is what they were they were talking about. Since they are also talking about semiconductor/metal transitions (rather than a near-black IR material), the materials may have a high index of refraction and therefore a significant reflectivity, and therefore having a second chance would be beneficial. $\endgroup$ – uhoh May 13 '17 at 13:05

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