# Skinsuit for a superman

Gold Member
I’m writing some stories about a superhero flying brick; similar to Superman in that he has superspeed as a power. He has a problem in that he has no indestructible suit like Supes employs. However, some friendly advanced aliens presented him with a skinsuit that can handle speeds (at sea level) of up to 4,000 mph before it heats up, etc, and disintegrates. It can handle up to 500 gees of acceleration also. But, Mr. Brick can also travel at up to better than half light speed (roughly 100,000 miles per second), so he can travel at handy interplanetary speeds.

In outer space, should this 500 gee max still hold for his skinsuit, to be reasonably consistent, physically speaking? If so, my calculator says it will take him 20+ days to accelerate to half c. That would put a cramp in his interplanetary escapades, and I'd have to do yet more handwavium. :)

## Answers and Replies

Related Writing and World Building News on Phys.org
In outer space, should this 500 gee max still hold for his skinsuit, to be reasonably consistent, physically speaking? If so, my calculator says it will take him 20+ days to accelerate to half c. That would put a cramp in his interplanetary escapades, and I'd have to do yet more handwavium. :)
You might want to invest in a better calculator. Can your storyline handle 8-9 hours to ramp up/down ?

Last edited:
chasrob
Staff Emeritus
2019 Award
Why only 500 g's? My hard disk will survive that if it's off. Don't see why a pair of trousers would be less robust.

chasrob
Gold Member
You might want to invest in a better calculator. Can your storyline handle 8-9 hours to ramp up/down ?
You’re right, I’m wrong, my calculator misled me again ;). So that means superguy has traveled around 2.5 billion kilometers when he hits 100,000 miles per sec, correct? That’s almost half-way to Pluto...

Say he wants to get to Mars at that rate; accelerate to half point, de-accelerate. I saw some equations that would calculate his time, given the acceleration and distance, but I can’t locate them now, dammit. You wouldn’t know what those equations are? I can’t seem to be able to Google the right words to describe what I’m looking for.

Gold Member
Why only 500 g's? My hard disk will survive that if it's off. Don't see why a pair of trousers would be less robust.
Good question. Off the top of my head, would 10,000 gees be more reasonable for a skinsuit that can handle 4,000 mph at sea level, you think?

hutchphd
Seems to me that is a question for the Aliens, assuming they have a head with a "top"........

chasrob
1d kinematic formulae : Sticky post in the Intro Homework forum. Enjoy.

I'd suggest working backwards, though : figure out how much time you want the superhero to spend in transit between various A-to-B's and work out the base figures for acceleration and max speed from that.

Toss in the occasional dust cloud, orbital-debris field, particle storm, or the like for filler : if you hit *anything at all* at 0.6c, you're going to need an advanced alien tailor.

Last edited:
chasrob
Gold Member
Seems to me that is a question for the Aliens, assuming they have a head with a "top"........
He goes 4,001 mph in atmosphere, the skinsuit disintegrates and he's left with his birthday suit :). These aliens are not going to give the crypto-earthling any more advanced tech that he can reverse engineer.

hutchphd
Gold Member
1d kinematic formulae : Sticky post in the Intro Homework forum. Enjoy.

I'd suggest working backwards, though : figure out how much time you want the superhero to spend in transit between various A-to-B's and work out the base figures for acceleration and max speed from that.

Toss in the occasional dust cloud, orbital-debris field, particle storm, or the like for filler : if you hit *anything at all* at 0.6c, you're going to need an advanced alien tailor.
Thanks. I do know about the SUVAT equations. I just remember, a couple years ago, coming across an equation or set of equations*that gave the best time when given acceleration and distance. Or maybe I mis-remember.

Like in the example above; say Mars is 140 million miles away and superguy starts accelerating at 500 gees. He would go half-way then de-accelerate at 500 gees, correct? In this case, he wouldn't reach anywhere near 100,000 miles a second before having to de-accelerate, right?
*edit: equations derived from SUVATs, I mean

Last edited:
Gold Member
Also, 8 or 9 hours of constant acceleration would be pretty boring, eh? :)

Well, if I plugged in the figures properly to this online calculator , the trip'd take 3½ hours'ish, and turnaround would be at maybe 190million mph.

Last edited:
chasrob
Gold Member
Well, if I plugged in the figures properly to this online calculator , the trip'd take 3½ hours'ish, and turnaround would be at maybe 190million mph.
Super, thanks!
So if Mr. Brick gets a call that the Mars colony is under attack, puts on his skinsuit, accelerates and then de-accelerates at 500 gees, 3 and 1/2 hours later he's on Mars? But the Bugs have already nuked the colony in the meantime. All because he's a prude :).
Edit: With no clothes to encumber him, he can accelerate to almost 100,000 miles per second in a matter of a few nanoseconds.

Last edited:
DaveC426913
Gold Member
Well, if I plugged in the figures properly to this online calculator , the trip'd take 3½ hours'ish, and turnaround would be at maybe 190million mph.
The turnaround time is non-trivial. That 190 million mph will be in a direction perpendicular to his journey. It won't make his trip shorter. His actual turnaround speed - relative to Earth - is what matters - and it will be zero, no matter how wide or narrow a loop he makes around Mars.

(There's a relevant nautical term called "Way Made Good" that defines the speed you are making relative to your destination(s). In other words, it is only the one-dimensional speed in a straight line between Earth and Mars that matters.

The total trip will be
Getting up to max positive speed at max positive accel +
(Coasting) +
Getting down to zero speed at max negative accel +
Getting up to max negative speed at max negative accel +
(Coasting) +
Getting down to zero speed at max positive accel.

Last edited:
chasrob
Gold Member
That 190 million mph will be in a direction perpendicular to his journey.

His actual turnaround speed (relative to Earth) will be zero.

(There's a relevant nautical term called "Way Made Good" that defines the speed you are making relative to your destinations. If he loops around Mars,)

The turnaround time is non-trivial. No matter how wide or narrow a loop he makes, he still has to decelerate from his top speed (relative to Earth) down to zero (relative to Earth), and then back up again.

The total trip will be
Getting up to max positive speed at max positive accel +
(Coasting) +
Getting down to zero speed at max negative accel +
Getting up to max negative speed at max negative accel +
(Coasting) +
Getting down to zero speed at max positive accel.
Yeah, that's what I was thinking. Say the distance to Mars is 225 million kilometers. He'd accelerate half that, 112.5 million. His max velocity (I get from the page) is 75 million miles per hour. No coasting period in this case.

Last edited:
DaveC426913
Gold Member
Yeah, that's what I was thinking. Say the distance to Mars is 225 million kilometers. He'd accelerate half that, 112.5 million. His max velocity (I get from the page) is 75 million miles per hour. No coasting period.
And don't forget that, as far as pure efficiency goes, there is no reason to stray from a straight line there and back. His most efficient turnaround action is to simply decelerate to zero. "Looping around" only lengthens the journey.

chasrob
mfb
Mentor
That 190 million mph will be in a direction perpendicular to his journey.
Huh? It's a one-dimensional motion problem.

Accelerate forward until you are halfway there, ~110 million km in the given example, then accelerate backwards. t=sqrt(2s/a) = 6600 seconds for each half of the trip, or about 3.5 hours total to get to Mars.
No need to make this any more complicated.

Top speed will be ta=33,000 km/s or 11% the speed of light.

Put the suit into a sturdy box and you can accelerate much faster.

chasrob
Gold Member
Huh? It's a one-dimensional motion problem.

Accelerate forward until you are halfway there, ~110 million km in the given example, then accelerate backwards. t=sqrt(2s/a) = 6600 seconds for each half of the trip, or about 3.5 hours total to get to Mars.
No need to make this any more complicated.

Top speed will be ta=33,000 km/s or 11% the speed of light.

Put the suit into a sturdy box and you can accelerate much faster.
The box would survive, but would the suit inside?

mfb
Mentor
Put a pair of trousers flat on the ground, stand on a random place on them. You probably stand on ~50-100 gram of clothing or so with a mass of ~100 kg. The force between trousers and ground is now equal to the force you get at ~1000-2000 g (without you standing on top). Do you risk any damage to the trousers? Probably not - they can withstand much larger forces. That's a random clothing item from your wardrobe. You would expect some alien super-material to be much sturdier.

When Mr. Brick wears the items they are subject to tension, but in a sturdy box you only get compression, which is much easier to handle.

chasrob
Gold Member
Put a pair of trousers flat on the ground, stand on a random place on them. You probably stand on ~50-100 gram of clothing or so with a mass of ~100 kg. The force between trousers and ground is now equal to the force you get at ~1000-2000 g (without you standing on top). Do you risk any damage to the trousers? Probably not - they can withstand much larger forces. That's a random clothing item from your wardrobe. You would expect some alien super-material to be much sturdier.

When Mr. Brick wears the items they are subject to tension, but in a sturdy box you only get compression, which is much easier to handle.
Yeah, you're right. That figure I just snatched out of thin air... 500 gees. Seemed to be reasonable since it's a tremendous force applied over a length of time. Maybe alien super-material could handle tens or hundreds of thousands of gees.

DaveC426913
Gold Member
Well, if I plugged in the figures properly to this online calculator , the trip'd take 3½ hours'ish, and turnaround would be at maybe 190million mph.
Huh? It's a one-dimensional motion problem.
Yes it is.

(Upon rereading, I may have misinterpreted what hmmmm27 meant by "turnaround" - I interpreted that to mean "looping around Mars" - and so went off on that tangent. Mea culpa)

DaveC426913
Gold Member
Yeah, you're right. That figure I just snatched out of thin air... 500 gees. Seemed to be reasonable since it's a tremendous force applied over a length of time. Maybe alien super-material could handle tens or hundreds of thousands of gees.
BTW, depending on context, 500gees is quite mundane.

A 10kg object, falling 1m onto a hard surface like concrete (~2ms duration), experiences something on the order of 500 gees.

(Upon rereading, I may have misinterpreted what hmmmm27 meant by "turnaround" - I interpreted that to mean "looping around Mars" - and so went off on that tangent. Mea culpa)
"turnaround point" as a mis-remembered phrase from an old novel (probably Heinlein, probably fusion-powered rockets), referring to a midpoint attitude reversal at full acceleration.

DaveC426913
Gold Member
"turnaround point" as a mis-remembered phrase from an old novel (probably Heinlein, probably fusion-powered rockets), referring to a midpoint attitude reversal at full acceleration.
Yes, of course.

I glommed onto the idea that it was a round-trip, so I thought of "turnaround" as the loop around Mars, and ran with it. How embarrassing.

Gold Member
BTW, depending on context, 500gees is quite mundane.

A 10kg object, falling 1m onto a hard surface like concrete (~2ms duration), experiences something on the order of 500 gees.
How about 500 gees applied over 3.5 hours, as in the above scenario? Cannot be very pleasant.

I glommed onto the idea that it was a round-trip, so I thought of "turnaround" as the loop around Mars, and ran with it. How embarrassing.
My fault : "turnaround time" is what it sounds like : the time it takes to refuel and reorient for the next or return leg of a trip.

What's wanted is a term for the midpoint turnover (an otherwise delightful fruity pastry) of a brachistochrone trajectory : Heinlein used the term "skew flip".

Anyways, as far as the suit is concerned : just fold, and store in a convenient orifice during those pesky interplanetary and "through a neutron star" trips.