Probability of Collison of Starship and stuff - stuff twixt earth and near star

starship1 · 2004-06-08 22:12:28

I am new to the group and want to know if my fuzzy math solution below is in the ballpark?

Probability of collision of my constant 1 g accelerating plasma rocket engine with space debris in journey to near star.

google search term...Poisson "probability of collision"......finding
quote..........

ON-ORBIT COLLISION HAZARD ANALYSIS IN LOW EARTH ORBIT
http://ast.faa.gov/files/pdf/poisson.pd … oisson.pdf

The equation one Poisson probability distribution function is given by the equation
Pk=-(R^k)(e^-k)/k!
P= probability of k events
k=number of events
R=rate of occurrence parameter

equation four (partially derived from Poisson)
PC =1 -exp (-AC*SPD*VR*T)
PC = Probability of collision for the duration of time T
AC=cross sectional area, km
SPD = Spatial density, objects/km
VR = relative velocity, km/s and
T=Time at risk seconds

Therefore making velocity faster reduces the time at risk and the smaller the cross section the better as is the lower the density. however I am getting a contradiction as higher velocity makes the number in parenthesis bigger instead of the expected smaller. So I do not use the equation with confidence at this time until I read more.
On-Orbit Collision Hazard Analysis in Low Earth Orbit Using the Poisson Probability Distribution (Version 1.0)

http://ntl.bts.gov/card_view.cfm?docid= … ?docid=219

Probability of Collision Error Analysis

http://jungfrau.tamu.edu/~html/alfriend … ...per.pdf

I could be only a partial derivative equation and I am misreading the exp notation as inverse natural log as sometimes exp notation works only in base 10 and the natural log is 2.7128 something base as I fuzzily recall.

Brushing up on my probability theory and log math, unless you want to take a stab at doing the probability collision calculation yourself with the values I give in this post unless you have some others you prefer

My gut feeling in any full derivative equation, I have to divide the length of trajectory (4.25 light years) by the length of the ship (approx 300 meters )getting a denominator under the nominator of 4.25 ly times ship width estimated at 6 meters 6 meters/10,000,000 objects to get the spatial density low requirement--if that is some help and you want to give it a try.

I have a exp button on my windows scientific calculator, but guess it just gives a the scientific notation of a number. I used the ln and inverse ln (log number) button recently with good results if I trust a calculator - How's your math? I have some reading to do from the google search term above.

....reading....Okie dokie I will take a stab at it now. Jumping in.....

exp is indeed the inverse natural log function (ln) as
exp(x)=1/ln(x) so to use my inverse ln button as I do not have a exp button

PC =1 -exp(-AC*SPD*VR*T) becomes
PC =1 - (1/ln(-AC*SPD*VR*T)

PC = Probability of collision for the duration of time T
AC=cross sectional area, km squared
SPD = Spatial density, objects/km cubed
VR = relative velocity, km/s and
T=Time at risk seconds

checking first physics units 1km^2 * 1/km^3 * 1km/s * 1sec = 1 so units are correct leaving a real number as dimensionless value.
AC = pi * radius of ship squared = 3.14 times (.003 km)^2 = 0.00002826 km^2
That was easy enough giving me confidence to figure the next three variables

Java calculator to figure trip time found at:
http://ucsu.colorado.edu/~obrian/applet … oyage.html also see http://www.fourmilab.ch/cship/craft.htm … craft.html

Trip length: 4.25 light years (ly)
Acceleration: 1.0 g wrt earth mid way to decelerate at one 1 g to arrive at
Near star at rest.......
Trip length: 4.25 light years (ly)
Time on earth: 5.8780560467144 years.
Time on ship: 3.544401860293398 years

As Velocity = distance traveled/ time traveled
Velocity average wrt earth = .732 C and c =299,792.458 km/s
Vavg=219448.079256 km/s

T = 5.8780560467144 years times 365.25 days/ year times 24 hours/day times 60 minute/hour times 60 seconds/ minute
= 185497341.49979434944 seconds as Time at risk

two easy down two to go.......

SPD = The Local Bubble extending out 500 light years is not only relatively empty (with a density of less than 0.001 atoms per cubic centimeter); it is also quite hot, about one million degrees Kelvin. By comparison, the interstellar cloud around the solar system is merely warm, about 7,000 degrees, with a density of about 0.3 atoms per cubic centimeter..........the relative sun/cloud velocity is 26 kilometers per second..........

citations for above paragraph-http://www.americanscientist.org/template/AssetDetail/assetid/21173/page/2
http://spaceflightnow.com/news/n0305/30 … 5/30map3d/

Picking a good number between .001 atoms/cc and .3 atoms/cc, I pick .003.cc or three times the known installer density accounting for a little of the higher density in the cloud around the sun averaged over the distance to the star multiplied by the higher velocity number applied to the solar cloud to gives a very conservative estimate

SPD = objects/km^3 = .003 atoms times1,000,000,000,000,000 cc =3000000000000 atoms/km^3
as the trajectory total volume = area times height
= 0.00002826 km^2 times 4.25 light years
Vol.tot = 0.00002826 km^2 times 299,792.458 km/s times 185497341.49979434944 sec
= 1571558493.9290642025410120746752 cubic km atoms = 3000000000000 atoms/km^3 times
1571558493.9290642025410120746752 /km^3
= 4714675481787192607623.0362240256 atoms crossing my path

side bar........how much mass?....
There are 6.0225x10^23 hydrogen atoms in 1 gram of hydrogen
so grams of hydrogen atoms in my path
= 0.0078284358352630844460324387281445 grams

So if I hit every single atom in my path at an average angle of
45 degrees
Ekinetic= (0.0078284358352630844460324387281445 grams times
216822.079256 km/sec times 216822.079256 km/sec)/2
= 184014484906.20450984639531873246 kilograms-m^2/sec^2
= 184014484906.2 watts over 5.8 years
= 0.99200604935035425532482688826712 watts per second

That amount of energy released in atom collisions my ship can withstand.
Note at .72 C relativistic energy kicks in so multiply that by approx
1.5 to guesstimate the gamma factor Ek=EC^2 *gamma/2 and
gamma =1/ square root(1-V^2/C^2) maybe? .72 times .72 ~= .5 ? I will look it up later

Back on topic- since I guess the velocity of the atoms is proportional to the temperature and I know the average velocity of the sun cloud atoms at 7,000 K, I can calculate the higher velocity of the interstellar atoms at 1 million K by multiplying the velocity of the solar cloud atoms by the ratio 1,000,000/7000 to get the higher velocity of the interstellar atoms

VR = velocity of ship - velocity of objects= (26 km/sec)(1,000,000K/7,000K)
=ship velocity - 3714.2857142857142857142857142857 km/sec ship velocity
= Trip length: 4.25 light years (ly) / Time on earth: 5.8780560467144 years.
Velocity = distance traveled/ time traveled
Velocity average wrt earth = .732 C and c =299,792.458 km/s
Vavg=219448.079256 km/s
VR = 219448.079256 km/s - 3714.2857km/sec
= 215733.79354171428571428571428572 km/sec

however as the objects move at various angles to the trajectory path I will
assume average of 45 degrees so correct the velocity by the
sine of 45 degrees = .707 and correcting
VR = 219448.079256 km/s - 2626 km/s = 216822.079256 km/sec

knowing now AC, SPD, VR and T; I solve for PC

PC =1 - (1/ln(-AC*SPD*VR*T)
= 1 - (((1/ln((-0.00002826 km^2 )(3000000000000/km^3)(219448.079256 km/sec)(185497341.49979434944 sec)))
= 1 - (1/ln(-3451142452668224988780.0625159867)

Note: ln(minus x) returned invalid input so I dropped it recalling that velocity is not a unit vector but scalar vector having both magnitude and direction with the direction noted by the minus sign so I should have subtracted the larger ship velocity from the objects smaller velocity getting a negative velocity to multiply the minus sign by making the ln(-x)= ln(x) if I recall my analytic geometry class from 1973. That solves the previous contradiction I found of a high V when expecting a low V as minus numbers are lower than positive numbers. Besides I could not conceive of a way to make time at risk a negative number or understand negative objects or imagine a negative density leaving only velocity to change and relative velocities are often negative
therefore..............

PC = 1- 1/49.592992274802526103814870677229
= 0.97983586079140287384216440679304 times 100 percent
= 97.9835860 percent probability I will hit at least one atom

Recalling my combinations and permutations probability math from 1968
The PC of n number of objects
=PC of one object raised to the nth power
= PC^n
Probability for hitting 10 atoms
PC^10 = 0.81570533089286596506609074604119 =81%
PC^100 = 13.041663959514299452836762753626 %
PC^1000 = .00000001423410145081292721735445731557 percent
considered insignificant if not zero therefore

of 1000 atoms hit /4714675481787192607623 atoms available = power of collision / power available of 184014484906.2 watts over 5.8 years
power of collision = 1000 atoms times 184014484906.2 watts
/ 4714675481787192607623 atoms
= 0.000000039030148653295137686211255652988 watts
also considered insignificant if not zero as the probability of releasing that energy in hitting 1000 atoms is considered insignificant if not zero.
care to check my math as I notice instead of The PC of n number of objects
=PC of one object raised to the nth power

it may actually be
Pk=-(R^k)(e^-k)/k!
P= probability of k events
k=number of events
R=rate of occurrence parameter

however just looking at some x/K! with out knowing x with K a thousand atoms its a very big number calculating at 4.02387260077093773543702433923e+2567 or 4 times ten to the 2567 power so Pk for 1000 atoms is also considered insignifigant if not zero.

How do you like my fuzzy math calculations?

It would seem the same equations can be used to protect the crew from alpha, beta and gamma radiation known to be present by simply moving a large quantity of the propellant water at high velocity in a hollow ship skin a few inches thick as a shield around the crew. Increasing the velocity of the water molecules would increase the probability to near 100% of those high velocity particles of postive, negative and neutral charge colliding with the water instead of the crew.
As the permissible radiation dose is 25 REM over the life time of the crew and current measurment of radiation in space is unknown by me.

For delicate electonics the dose is http://www.hq.nasa.gov/office/pao/Histo … /app-1.htm
"For space applications, the recommended dose rates should be kept as low as possible, preferably 0.01 - 2 rads(Si)/sec.
care to to plug in other numbers and do that math anyone?

Mundaka · 2004-06-09 01:11:36

bolbuyk · 2004-06-09 02:28:47

Interesting stuff, but something s going wrong in your calculations.
In the pdf-file, éxp' means e^, so your formula is:

P=1-e^(-AC*SPD*VR*T). This nis more logical, because from the former formula is derived, as is rather clearly explained in the pdf-file, that P(0)=e^-R, so P(>0)=1-e^-R. The value of e^-R is always smaller than 1, because R is always positive. In case of logs, this doesn't work.

The T can be derived from the VR and the distance that is traveled (D). T=D/VR (D is not the shortest line between the two points burt the real distance that is traveled). So the formula becomes:

P=1-e^(-AC*SPD*D)

In other words: Imagine a tube with the length of the distance and the cross-section of SPD. If there are more objects in this tube, the probability of collission increases. The time and velocity are ruled out.

Palomar · 2004-06-09 09:44:13

Well, uh . . . Hey! How about those Mets!?!

:laugh:

***
Well, I'm not very good with math (understatement).

On the optimistic (and non-mathemetical) side, we've not had many problems with probes being struck by space debris; even Apollo 13 wasn't hit by debris. IIRC, one of Japan's probes got hit, but none other than I can recall off-hand. I wonder if the probes are beating the odds!

A space-faring vessel ala Mars Direct would be larger than a "regular" probe...and of course the bigger the vessel the greater the chances of getting hit by something.

The success of our long-faring probes (Voyager series especially), etc., may hold out some additional optimism.

Not sure if this helps in any way... :-\ And sorry I can't assist with math.

--Cindy

starship1 · 2004-06-09 18:48:01

Interesting stuff, but something s going wrong in your calculations.
In the pdf-file, éxp' means e^, so your formula is:
P=1-e^(-AC*SPD*VR*T). This nis more logical, because from the former formula is derived, as is rather clearly explained in the pdf-file, that P(0)=e^-R, so P(>0)=1-e^-R. The value of e^-R is always smaller than 1, because R is always positive. In case of logs, this doesn't work.
The T can be derived from the VR and the distance that is traveled (D). T=D/VR (D is not the shortest line between the two points burt the real distance that is traveled). So the formula becomes:
P=1-e^(-AC*SPD*D)
In other words: Imagine a tube with the length of the distance and the cross-section of SPD. If there are more objects in this tube, the probability of collission increases. The time and velocity are ruled out.

very good, the term VR times T is replaceable with distance as you point out T=D/VR and cross multiplying both sides by VR then D=(T)(VR) simplfying the equation and this passes the physical units tests as 1 meters^2 times 1/(meters^3) times 1 meter = 1 so the answer is indeed a real number of no units so dimensionless and valid as an exponent;however the number AC*SPD*VR*T is the same as AC*SPD*D so my answer remains the same and as distance traveled is derived from the velocity and time than both are still relevant and both are valid equations. I think that I will keep the Vr times t factor as both velocity and time in physics are measurements so though it simplfies the math to factor at this point I loose the error magnitude in both variables so it may not serve me in the long run to wait and factor before the final calculation exspecailly considering a constant 1 g accelerating object as it has a max velocity and most of the distance covered is at far above average velocity reducing the time at risk lost and data lost if you just consider the distance traveled. It may be more by packrat habit as I do tend to carry all the data I can till I must discard it.

I imagine 10,000,000 10 pound rocks crossing my cylindercal path 4.25 light years long at various angles averaging 45 degrees. The probability of any rock being in the same space and time that my rocket is during the trip time drops as velocity increases reducing the time at risk therefore I would expect the probability of collision to drop to a value considered insignifigant if not zero, even in the unlikey case of 10,000,000 rocks.

That leaves the minus sign still in question as you point out. No having a exp button on my calculator and knowing that ln was the inverse function of exp I derived the equation that may have been incorrect as I misinterpeted the minus sign inverting it..

quote...........
http://spicerack.unh.edu/mac/calc/power … power.html
The logarithmic function that arises naturally out of calculus is called ''ln.'' Its name is pronounced ''natural log'' or sometimes just ''log'' or by sounding out its spelling: ''el-en.''
Its inverse exponential function is ''exp'' and its value-variable is called either ''exp(x)'' or ''ex.'' You can pronounce it either as ''exp'' or ''e-to-the-x.'' The first notation, exp(x), reminds us that the number x is an argument to the function: the second notation, ex, reminds us of the rules of exponents that the function's values follow..............

All the ''rules'' of logarithms are the same, whether we use 10 as a base (common logarithms) or e as a base (natural logarithms). The systems are even proportional!

e^lnx = x = ln(e^x)
end quote ............

So now I am totally confused about the given minus sign or if I gave the correct inversion idenity above to arrive at my answer perhaps you can plug in your ac, spd and d values but using the exp function to check my answer htta may be correct by accident as I still cannot do exp funtions on my windows scientific calculator. One may be able to do them but I just have not figured it out.

I thought another had it figured out from the dynamic rocket equation
Vf = Ve * LN(1/(1-u))
from which the inverse identity is
u = 1 - 1/EXP(Vf/Ve)
with u being the propellant fraction of final Mass/ initial mass mass or initial velocity over final velocity conserving momentum but apparently not so if you can give me clue and
solve for
PC = 1-exp(-AC*SPD*VR*T)
PC=1-exp(-3451142452668224988780.0625159867)

to see if it returns the fuzzy value below

PC =1 - (1/ln(-AC*SPD*VR*T)
= 1 - (((1/ln((-0.00002826 km^2 )(3000000000000/km^3)(219448.079256 km/sec)(185497341.49979434944 sec)))
= 1 - (1/ln(-3451142452668224988780.0625159867)
PC = 1- 1/49.592992274802526103814870677229
= 0.97983586079140287384216440679304 ???

bolbuyk · 2004-06-10 05:25:09

When you look at the pdf-file, it is clear that they means by 'exp' e^ . The minus sign is logical.

The formula: P=1-e^(-AC*SPD*D)

The bigger the amount of stuff, the longer the distance, the bigger the surface, the more negative the value becomes. So e remains always smaller than 1 and bigger than 0. The more negative the value, the more e^ comes to 0. 1- then gives a chance that comes close to 1.

Most times, as far as I know, EXP is used for 10^or ln, but in this case it is obvious it is used for e^ . confusing they use the latter notation in their first formula and the EXP in the latter.

Cindy: As far as I know just two manned spacecraft were once hit by some stuff: The Salyut 7 and the Space-Shuttle (don't know which mission). In both cases it gave no serious problems, just a 'big bang' and some small damage.

Problems like this only become important when travellibng over very large distances (or circling many many rounds around a planet).

MarsDog · 2004-06-10 08:45:51

An article proposed using, space based, high powered lasers to clear a path within the solar system.
-
In interstellar space, possibly use a laser on low power as a radar, then high power to blast the target.

starship1 · 2004-06-10 14:00:34

When you look at the pdf-file, it is clear that they means by 'exp' e^ . The minus sign is logical.
The formula: P=1-e^(-AC*SPD*D)
The bigger the amount of stuff, the longer the distance, the bigger the surface, the more negative the value becomes. So e remains always smaller than 1 and bigger than 0. The more negative the value, the more e^ comes to 0. 1- then gives a chance that comes close to 1.
Most times, as far as I know, EXP is used for 10^or ln, but in this case it is obvious it is used for e^ . confusing they use the latter notation in their first formula and the EXP in the latter.
Cindy: As far as I know just two manned spacecraft were once hit by some stuff: The Salyut 7 and the Space-Shuttle (don't know which mission). In both cases it gave no serious problems, just a 'big bang' and some small damage.
Problems like this only become important when travellibng over very large distances (or circling many many rounds around a planet).

PC = 1-exp(-AC*SPD*VR*T)
PC=1-exp(-3451142452668224988780.0625159867)
PC= ?

What I asked is for someone to actually calculate that above value for PC with that given number and filling in the question mark as I do not know "how to" on my windows scientific calculator. I can then check it with value of the inverse log I calculated and if different I will change my equations as my derivitive was incorrect.

By experiment, simply running across the street to a doorway during a rain shower I get less wet than if I just mosey across the street as my increased velocity reduces the probability of collision with rain drops as it reduces my time at risk.

I do not know anything about the mets. Are they a football team:), but I still enjoy all the readers comments. A laser is not a bad idea but if I can hit the object with a laser, then I can probably just steer around it easier.

It seems from the below links that something has already swept this part of the galaxy clean leaving a clear path to most all near stars.

They even have photo of the path
http://www.americanscientist.org/templa … mplat....2
http://spaceflightnow.com/news/n0305/30 … 5/30map3d/

MarsDog · 2004-06-10 14:49:55

Looks like safer trip to the nearby stars.
The meteor fields of science fiction might have also been moved by the blast. Too bad we won't be alive to see the results of interstellar probes. We need bigger telescopes.

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#1 2004-06-08 22:12:28

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#2 2004-06-09 01:11:36

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#3 2004-06-09 02:28:47

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#4 2004-06-09 09:44:13

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#5 2004-06-09 18:48:01

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#6 2004-06-10 05:25:09

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#7 2004-06-10 08:45:51

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#8 2004-06-10 14:00:34

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

#9 2004-06-10 14:49:55

Re: Probability of Collison of Starship and stuff - stuff twixt earth and near star

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