All you have to do is plug it in!
Posted on 05/10/2022 6:51:02 PM PDT by LibWhacker
All you have to do is plug it in!
Spinlaunch gets package to 200,000 feet where rocket takes over and accelerates it to 17,500 mph.
Yeah, but can you thread a needle?
Sorry, no: If the hoped-for "muzzle velocity" (i.e., velocity of the payload upon exiting the centrifuge) is 5,000 mph - as stated in the article - then it will never be able to reach orbit. It could, at best, travel a few hundred miles before crashing back onto the Earth's surface.
This is even NOT taking into account such factors as air friction.
Fail!
BTW: I don't actually fully understand your statement. What do you mean by initial velocity and/or how does it differ from launch speed?
Regards,
No mention in the excerpt of any rocket to assist in attaining orbital velocity - should have been highlighted in the FIRST PARAGRAPH - but that is the only scenario I can imagine having even a chance of working.
There are artillery shells that employ rocket-assist - but the g-forces involved in centrifuging something to 5,000 mph would be enormous.
And I'd also like to know what kind of payloads we're talking about here. Few things could survive such g-forces.
Regards,
The Earth's gravity drops only insignificantly until you are many hundreds of miles above the surface, in space.
The decrease can be safely ignored when performing "back-of-the-envelope" calculations pertaining to low-earth orbit.
Regards,
You are truly a man ahead of your time!
BTW: Where did you get your degree in Physics?
Regards,
Initial velocity and launch speed are same thing in my post. Did not word things very well. In any case, this process supposedly would perform the necessary acceleration at ground level. I am curious as to whether that makes a difference in how much speed must be attained to remain in orbit. Same goes for trajectory: does it effect velocity bottom line? (I doubt it.) As for friction at time of lauch, I suppose they could apply countermeasures akin to what we’ve done with inbound craft, hamely a heat shield.
Well, they do launch rockets to the East, in the direction of earth’s rotation, to pick up that extra 1,000 mph.
I’ll always remember the derivation:
Orbital velocity, F(gravity) = F(centripetal)
GMm/r^2 = mv^2/r; v^2 = GM/r; Vo = sqrt(GM/r).
Escape velocity, PE(gravity) = KE(velocity)
GMm/r = mv^2/2; v^2 = 2GM/r; Ve = sqrt(2GM/r).
Ve = sqrt(2)Vo.
Right. But take a look at the video. The sabot is apparently made of some kind of graphite, and the velocity of the projectile is not supposed to be orbital. The “throw” only gets the interior rocket minus the sabot up high enough and fast enough to fire at an altitude saving LOTS of fuel. I think this is a winner.
A theory well worth testing.
Okay.
In any case, this process supposedly would perform the necessary acceleration at ground level. I am curious as to whether that makes a difference in how much speed must be attained to remain in orbit.
Okay, let's first make it clear that, in this scenario, we are talking about a one-time, brief acceleration imparted to the object we want to place into Earth orbit - like from a cannon, as opposed to a rocket which slowly accelerates the payload.
Aside: Even the Space Shuttle - which had human passengers, thus making it necessary to only slowly accelerate over a period of many minutes - sustained g-forces of 5 to 6 were unavoidable during a certain phase of the launch.
In the context of such ballistic launches, the expression "orbital velocity" then means "the velocity which, when imparted to an object in a very short period of time, will suffice to place it into a stable orbit (i.e., not just a sub-orbit - meaning that it would soon fall back to Earth - but rather into a true orbit, meaning that it will remain more or less indefinitely in space)."
The payload is thus basically a projectile that is brought from zero to 18,000 mph in an extremely short span of time.
The least amount of energy needs to be imparted to the projectile when it is accelerated parallel to the Earth's surface. Of course, because of mountains, tall buildings, and especially Earth's atmosphere, that is not practical. (In the case of a smooth, airless celestial body, the stable orbit could be a few meters above ground level!)
But the orbital velocity would have to be only slightly higher in order to place the payload into an orbit at, say, 400 km above sea-level (approx. the altitude of the ISS).
Same goes for trajectory: does it effect velocity bottom line? (I doubt it.)
Yes, the trajectory DOES affect the "bottom line."
As I said above: The energetically most-favorable trajectory is parallel to the Earth's surface. Because of the unevenness of the Earth's terrain, the trajectory would have to be at a slight angle (a few degrees).
For practical reasons (air friction / drag), it's best to choose a trajectory that gets you out of the lower atmosphere as quickly as possible. (Solving that problem would require some university analytical calculus which I have long since forgotten.) So the optimal trajectory might depart from the parallel by a few more degrees.
As for friction at time of lau[n]ch, I suppose they could apply countermeasures akin to what we’ve done with inbound craft, [n]amely a heat shield.
The air friction results in two major problems: Drag (compensated for by adjusting the trajectory slightly upward - but also by having to increase muzzle velocity, since the drag is going to "devour" some - perhaps an APPRECIABLE FRACTION - of your initial velocity before you leave the lower atmosphere behind you) and thermal effects.
The Space Shuttle's thermal tiles protected it against a couple of minutes of extreme heat experienced when re-entering and decelerating at a piddling 1-2 g, going from roughly 18,000 mph to 0 over the course of a few minutes.
I don't know what kind of shielding - if ever - could be developed to protect what is essentially a shooting star in reverse.
I suspect that only some form of ablative armor could be used - thus further increasing the mass of the projectile.
Regards,
My understanding is that the centrifugal aspect of this device allows for gradual accelleration, not instant velocity like that produced by a cannon. Otherwise a good part of the payload would be subject to loss at launch. In view of the fact they are planning a larger devuce like this one, the question would be what diameter and rpm could best achieve the desired speed at launch time.
You’ve been more than kind and thorough in your replies. Thank you.
Sorry that I was unclear: The fact that the centrifuge only slowly increases in rpm's is of NO ACCOUNT.
In fact, that only ADDS to the time that the payload has to endure high g-forces.
The fact remains that, just prior to release (launch), the payload is being spun thousands of times per second, and is being subjected to incredible g-forces!
This could be likened to heating your body up to 140 °F: Would you rather I put you in a bath and slowly increase the temperature (for, say, an hour) until 140 °F is reached? Or would you rather I plunge you into a bath already at 140 °F?
After release, the projectile would then immediately begin decelerating as drag does its thing.
From, say, 20,000 mph to 18,000 mph in two seconds. This would (again!) result in fantastic mechanical stress!
I can see this whole concept working only if there are a number of additional considerations which the excerpt isn't mentioning (Star Trek-type super-strong constructional materials, etc.).
Less fantastic aids: The centrifuge could (perhaps) be carried to a high altitude by balloon - say, to above 90% of the Earth's atmosphere. A high mountains might also be feasible. The projectile would have to be equipped for rocket-assist (it would thus not be truly ballistic). But even then, I think that it would be practical only for bulk goods, like water. I can't imagine any conventional manufactured items surviving!
The whole thing is an engineering nightmare!
Regards,
In fact, that only ADDS to the time that the payload has to endure high g-forces.
Ican be thick of skull at times. It just occurred to me that g-forces would be in play on at least one axis during rotation and accelleration, no matter what length of time, and increasing all the more with speed. Whatever the payload, human occupation would be out of the question. The length of time necessary for heat shield may be comprable to, or less than, that on the shuttles of late, no?
No, as I already explained, the projectile would have a "muzzle velocity" (for want of a better word) of, say, 20,000 mph - meaning that, if launched at sea level, it would leave roughly 90% of the Earth's atmosphere behind (below) it in approx. 2-3 seconds. Or make that 24,000 mph and TEN seconds, if the trajectory is, say, 15° above the horizontal. (I am unable to calculate the effect of air friction and so am making ballpark estimates).
The Space Shuttle's ceramic thermal tiles had to withstand the heat of re-entry for about TWELVE MINUTES, as its speed gradually dropped from 18,000 mph to the speed of a fighter jet BY THE TIME IT REACHED THE DENSER LAYERS OF OUR ATMOSPHERE.
This centrifuged projectile, in contrast, would BEGIN in the deadly, denser layers! It would have to plough through our dense troposphere and stratosphere at more than 18,000 mph! (Probably at much more, to compensate for the loss of speed due to drag - its FINAL speed when entering space must be at least 18,000 mph, after all!)
Bear in mind, after all, that most "shooting stars" burn up in our MESOSPHERE - which approximates a good, hard vacuum!
This centrifuged projectile would therefore have to endure unbelievable g-forces, then unbelievable (higher than for the Space Shuttle) temperatures (albeit only for a few seconds) and again high negative g-forces (as it decelerated).
Regards,
Fester Chugabrew: The amount of g-force would be negligible if the rate of acceleration were low enough.
I can make neither hide nor hair of that statement.
The centrifuged projectile must be brought, within the confines of the centrifuge, up to orbital velocity (plus X to account for drag).
You could bring it up to speed either gradually or very quickly. Wouldn't matter: The final velocity must be more than 18,000 mph.
All that time, the g-forces are increasing - either gradually, or very quickly.
But at the moment in time of release (when the muzzle velocity is 18,000 mph plus "X"), the g-forces would be the same, regardless. They would be FANTASTICALLY HIGH!
Moreover, if the trajectory were in accord with/parallel as much as possible to the earths rotation, the need for speed would be lessened significantly, no?
No, not significantly. Measurably (roughly 5%), but not significantly - if you launched from the Earth's equator, towards the East.
Regards,
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