Impure Math

[bcholmes]

Math help! Is this right?

T2 = 2xD / A

Where T is time in seconds, D is distance in metres and A is acceleration in metres per second. I want to calculate the number of days it takes to travel X number of Astronomical Units -- AU (or ua, if you actually listen to the Bureau International des Poids et Mesures) -- using Gs of acceleration.

T2 = 2xD / 10xAG

Thus:

T2= 2xDAUx150,000,000x1000 / 10xAG

And:

Td2 = 2xDAUx150,000,000x1000 / 10xAGx(24x60x60)2

Which is:

Td2 = 2xDAUx150,000,000x1000 / 10xAGx86,4002

Or:

Td2 = 2xDAUx150,000,000x1000 / AGx74,649,600,000

Which is close enough to:

Td2 = 2xDAUx2 / AG

Or

Td = 2xSQRT(DAU / AG)

So at a constant one G acceleration I can travel 100 AU in about 7 days? That seems off, somehow.

Now, thanks to james_nicoll and king_tirian, I've got it right.

Comments

Isn't it S = 1/2 AT^2?

[james-nicoll.livejournal.com]

I get 100 AU in about 20 days or 28 if you need to slow down to rest at the end on account of how hitting things at 6% of the speed of light tends to upset the passengers.

The catch is looking at the implications behind being able to do this. See my recent LJ entry I Will Fly Torchship No More Forever for some of them (Like being able to fit a power plant with the output of Pickering into a juice jug). Avoiding having the ship vapourise as soon as you turn it on becomes something of a challenge with high accelerations combined with high exhaust velocities.

Re: Isn't it S = 1/2 AT^2?

[bcholmes.livejournal.com]

D'oh! Yes, thank you.

[king-tirian.livejournal.com]

I think all of your multiplication factors are on the wrong side of the equation. T^2 = (D_AU x 150,000,000x1000) / (AG x 10) in the third equation down, frex. In the end, it just means that the 2 in the denominator of the final equation should be in the numerator.

But I don't know if a 14 day trip makes you any less uneasy.

[cielf.livejournal.com]

I get more like 14 when I figure it out for 100 AU from scratch.

If what you want is a formula to determine the number of days to travel a specific number of AUs, at 1G -- I think the best approach might be to determine what 1G is in terms of AUs instead of metres... (using the very rough numbers, I get 6.7 * 10**-11.)

So T (in days) = t (in seconds) / 86400
= sqrt ( number of AUs / g (in AUs/sec**2))/86400
= sqrt (number of AUs / (6.7 * 10**-11)) / 86400

which still gives about 14 for 100 AUs. About 24 AUs in the first week. It takes almost a day and a half for the first AU...

Still seems like too little.

[kalikanzara.livejournal.com]

Do you want to travel it and stop, or pass it by? As in, do you need to count deceleration time as well?

[androgyna.livejournal.com]

Assuming uniform linear motion with constant acceleration, 1 g = 9.8 m/s², 1 AU (http://www.jas.org.jo/data.html) =1.49597870 x 10¹¹m, and an initial velocity of zero, I get the following approximate formula:

T_d = 2.02 x SQRT(D_AU/A_g)

where D_AU is the distanced traveled in AU, A_g is the acceleration in earth g's, and T_d is the elapsed time in days.

So your formula appears correct: i.e., I did the calculation independently.

TTFN
Michelle... ;)