# Unobtanium? **URL:** \ **Category:** Factual Questions\ **Created:** [June 18, 2014, 8:45pm UTC](https://boards.straightdope.com/t/unobtanium/691097 "2014-06-18T20:45:20Z")\ **Posts on this page:** 2\ **Page:** 3
**Author:** ![Habeed](https://avatars.discourse-cdn.com/v4/letter/h/a587f6/32.png) [@Habeed](https://boards.straightdope.com/u/Habeed)\ **Post date:** [June 28, 2014, 8:34pm UTC](https://boards.straightdope.com/t/unobtanium/691097/41 "2014-06-28T20:34:28Z")
> [@Chronos](#): > > 50% efficiency from a cannon? That’s ludicrously optimistic. Let’s say that we’ve got a cannon launching straight up, firing projectiles 10 m long, with a density about equal to water. The downward force from gravity on such a projectile is rho\_p_A_L_g, or a force per cross sectional area of about 100 kPa. The downward force from air resistance, meanwhile, is 0.5_C_rho\_a_A\*v^2. Without futzing about with Reynolds numbers, let’s just say that the drag coefficient is of order 1. Launch speed is going to be something comparable to escape speed, 10^4 m/s, and air density is about 1 kg/m^3. This gives us a force per area from air resistance of order 10^8 kPa, or a thousand times greater than the gravitational force. And it gets much worse if you’re launching at the very shallow angle a long-barreled cannon would need. > > Yes, there are some approximations I made there, but they work both ways, and the larger approximations are the ones that would make it even worse if accounted for. And yes, you can decrease the drag coefficient or increase the projectile length, but by three orders of magnitude? You only have forces in your math here. I don’t see any attempt to calculate the actual nonconservative losses from friction. Here’s where my 50% “fermi estimate” comes in. Your projectile will only be subject to those forces for a brief period of time. 5.5 kilometers is the point at which the atmospheric pressure halves. Since that air friction force is directly proportional to density (or pressure), if we assume velocity remains constant for the sake of an estimate, then all of the losses from 0 to 5.5 km = losses from 5.5 km to hard vacuum. Orbital velocity is a bit more than 5.5 kilometers/second. So you’ll reach 5.5 kilometers in maybe half a second. Also, your EM track is probably located on a tower on a mountain so that you exit the track several kilometers up. Technically, if subject to 1000 times the force of gravity for 1 second, you’d lose 10 kilometers/second in velocity. That sounds a bit much. I suspect you’re using the wrong equations. Also, that force would kill anyone aboard - I thought this type of cannon was at least possible for humans to survive if the track were long enough. If you’re right, then your speed exiting the cannon would need to be something on the order of 20 kilometers/second. That would make any plan to build a space gun fail the pencil test, and there’s a half dozen designs that pass basic analysis. I’ll have to look into this further. Losses in the superconducting cannon itself are minimal. This type of cannon has a 100% theoretical efficiency. (so it is reasonable to estimate the real cannon is 90% efficient) ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 29, 2014, 12:48am UTC](https://boards.straightdope.com/t/unobtanium/691097/42 "2014-06-29T00:48:22Z")
> [@](#): > > Losses in the superconducting cannon itself are minimal. This type of cannon has a 100% theoretical efficiency. (so it is reasonable to estimate the real cannon is 90% efficient) Yeah, that part I have no problem with. You will still have losses: Whatever you’re using to keep the projectile from rubbing against the walls won’t be perfectly smooth, you’ll need to maintain cryogenics for the superconductors, and so on, but the sum total is relatively small. There’s also a practical problem in how you transition from vacuum in the barrel (which would really _have_ to be evacuated) to atmosphere outside, but that wouldn’t cost much energy, just complication. [Previous page](https://boards.straightdope.com/t/unobtanium/691097.md?page=2)