mathletes, quick check:
Assuming constant acceleration of 0.1 G, zero initial velocity ...
Do I wind up with t = sqrt (D / a), where t is in seconds, D is distance travelled in meters, a is constant acceleration in meters per second per second?
Then, assuming I am travelling 0.5 AU (74.8 billion meters) before turnaround at an acceleration of 0.1 G (0.981 m/s^2), does it take me ~3.2 days (~276,000 seconds)?
Such that to travel 1 AU at 0.1 G constant acceleration, with turn around in the middle and arriving having shed my velocity takes ~6.4 days?
Is that about right?
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I haven't double checked your arithmetic but your process looks right
ReplyDeleteJust off the top of my head, I recall D as being 1/2 at^2, so ... your time factor seems like maybe it's off by a factor of sqrt-2?
ReplyDeleteiirc D=at^2/2 therefore t=sqrt(2D/a)
ReplyDeleteOf course it is!
ReplyDeleteBecause position is the second integral of acceleration, so as t is squared there's a one half in front.
So then 4.5 days to go 0.5 AU at 0.1 G, and nine days to go a full AU after turnaround.
ReplyDeleteThanks, Newton.
I disagree with Tony on this. Distance is the 2nd derivative of acceleration (acceleration -> velocity -> displacement). The thing is using calculus here just over complicates. You can just use unit analysis. D/a means you have a meters on top and a meters on bottom, which will cancel out. That leaves us with a sec squared under two divisions, which will flip it back to multiplication. So that leaves us with 1*sec^2. You took the sqrt to correct that, so as long as we don't have any negative numbers your good
ReplyDeleteThe problem I always have is that linear maths doesn't really cover what spaceships in a solar system need to be doing to move from one planet to another, while rotational physics to calculate required changes in angular momentum is that bit more complicated.
ReplyDeleteDavid Rothfeder The distance equation is a product of the calculus:
ReplyDeleteP(t) = P(o) + V(o) * t + 0.5 * a * t^2.
Just because the units cancel doesn't mean its right. That one half is essential.
LOL, this reads like a Peanuts teacher post to me.
ReplyDeleteCalculus not your domain of expertise, Mo Jave?
ReplyDeleteDavid: Man, that's hilarious. You can "disagree" with me about whether the equation for distance as a function of acceleration and time is (or is not) D = 1/2 a t^2, but don't expect much in the way of response. You don't have a valid alternate opinion, you're just wrong.
ReplyDeleteBrian Ashford That's for sure true. Especially as I'm assuming constant acceleration, no gravitation forces, completely ignoring relativity, etc etc.
ReplyDeleteCan't fly a spaceship that way, but I can sure do first pass approximations.
Hells to the no. I have mild dyscalculia. I can math, but not in the proscribed way.
ReplyDeleteMua-mua d-Mua/d-Mua.
ReplyDeleteTagging Brad D
ReplyDeleteTony Lower-Basch, scientists and mathematicians argue about proper methods all the time. They then provide proofs and write papers about how they're right. A method is accepted as fact when the evidence indicates one side is right. I may have been mistaken (I haven't done physics since I was about 22) but at the time you provided no proof so it was valid for me to disagree until you demonstrate your case. You being dismissive is in poor spirit and does little to encourage others to listen. If you can't act reasonably when somebody is mistaken then there is no use of even being right.
ReplyDeleteI'm just going to leave this here:
ReplyDeleteprojectrho.com - Mission Table - Atomic Rockets
hooray charts!
ReplyDeleteThat is a good chart.
ReplyDeleteDavid: Well, there's some use to being right... I helped William arrive at the right answer.
ReplyDeleteAnd, y'know, I've also apparently helped to create a wonderful case study in the Dunning-Kruger effect, so that's probably got some educational and entertainment value, as well.
I feel like you haven't demonstrated your case, on this assertion that being right is of no use.
Tony Lower-Basch, David Rothfeder If you're both having fun sniping, continue. If either or both of you isn't having fun, then don't respond. There's no points to arguing without it being fun.
ReplyDelete::popcorn::
I'm having a ball! Thanks for the heat-check, William.
ReplyDeleteNot having fun. Was offended as to how Tony was trying to correct a mistake I made. Was calling him out on it while attempting to remain proffesional.
ReplyDeleteWilliam Nichols: Over to you, man.
ReplyDeleteI'll just add this: Calling out is less likely to have positive outcomes than calling in. I recommend trying it next time, David.
ReplyDeleteLook, I was trying to say why his behavior was inappropriate rather than just saying "stop being an asshole". Do you think it's right to blame a person for keeping their cool after being insulted?
ReplyDeleteI think calling out is rarely helpful and leads to situations like this. That's why I recommend calling in.
ReplyDeleteOK, I was pretty calm about this before, but seriously? I was insulted. I told Tony why that was not right in a calm manner and your trying to tell me that I did so wrong to do so by spouting off a cliche? I've heard you do way worse. I've also seen you post about how gaslighting is a problem. I am not calm anymore. This might have gotten completely derailed from the original topic, but this pretty damned messed up.
ReplyDeleteAnd this is where I opt out of doing the invisible emotional labor to solve the problem. I can see it, and it won't be fun.
ReplyDeleteIf you're not having fun, then just walk away. There's plenty of places on the internet besides my front porch. Nobody wants you to be mad, but being mad is a decision you make -- nobody is making you be mad, or post agree, or perceive an insult.
and now, as this is my front porch and I got what I needed, I'm closing up comments.