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*To*: "'LIT Starship Design Group'" <starship-design@lists.uoregon.edu>*Subject*: RE: starship-design: Deceleration scheme*From*: "L. Parker" <lparker@cacaphony.net>*Date*: Tue, 5 Aug 1997 22:24:03 -0500*Reply-To*: "L. Parker" <lparker@cacaphony.net>*Sender*: owner-starship-design

Timothy, > I wonder if we can keep the sail from being ripped apart at more than 15g. Depending on areal density we can *generate* up to 700 g with a sail. It may take diamond filament tethers to hold on to it, but we can do it. > Furthermore the heating will likely melt away the sail. For a 99% reflective sail during a powered perihelion maneuver, the highest temperature we would expect would be only 1000K, well within the limits of many materials we have considered. Unless you REALLY stack up a lot of masers or lasers coming out of the solar proximity, the temperature should never exceed this. Of course, with lower reflectivity sails this temperature goes up. > I think we never came up with any number. We figured that there was just too > little to scoop to accelerate as fast as we wanted too. Grrrr...now you've done it...}:-| I HATE ASCII MATH! > OK, now you've done it. You've made me mad. >-| > > Sail size = A square m > Sail mass: M = A * 0.0001 kg/m^2 > > gamma at 0.9c = 2.29 >Momentum of the sail at 0.9c = gamma*M*v = A*0.0001*3E8*2.29 = A * 6.87E4 kgm/s Lots deleted for brevity... > It then will take (A*6.87E4)/(2*0.00423*A) = 8.1E6 seconds to stop the sail > This is about 100 days. What, you think light pressure from a star just "magically" cuts off at some distance from the star? Time for a refresher course: Light pressure on a reflecting sail is of course a function of distance from the Sun, but it also depends on how reflective the sail is. A completely absorbing sail (a "black body" sail) with light incident perpendicular to its surface of intensity, I (watts/ sq. meter), experiences a pressure, Ps, of magnitude: I Ps = --- C where c is the velocity of light. If the sail is 100% reflective, then: 2I Ps = ---- C Near the Earth, I is about 1400 watts/m^2. The solar radiation pressure on a thin piece of aluminized mylar could produce a force several times the gravitational attraction of the Sun on that piece of plastic. For sail reflectivity other than 100%, k between 0 and 1.0, the formula for pressure on the sail material is: Ps = (1 + k)I/c So if a sail is 95% reflective, the solar radiation pressure will be reduced by about 2.5%. Because the Sun's light decreases in intensity as it spreads outward, it will be reduced according to the inverse square of the distance, r, from the Sun - that is, by 1/r^2. So the pressure varies with distance as: (I + k)Io Ps = --------- r^2 c where Io is the solar light intensity at the surface of the sun. (The formula is reasonably accurate only for r> > Sun's radius.) Elementary units of light, photons, carry momentum as well as. The momentum, P, of a solar photon is equivalent to its energy divided by the speed of light: P = E/c. At the dawn of the space age, T. C. Tsu analyzed interplanetary solar sailing, using the following relationship for light pressure on a 100% reflective sail: 2Sr Pr = ---- c where Sr is the solar irradiance. Applying the inverse-square law, we demonstrated (7): Sr = (3.04 x 10^25)/r^2 watt/m^2 where r is the separation (meters) between the sail and the Sun's center. For a sail that does not transmit any sunlight though which has a reflectivity, k, we showed: (1 + k)Sr Pr = ---------- c If the combined sail, cable, and payload mass is M and the circular sail's radius is Rs, the acceleration of the sail outward from the Sun is: (1 + k) (6.3 x 10^17) Rs^2 as = ------------------------- m/sec^2 2Mr^2 The final interstellar departure velocity, V@, in meters/sec of a solar sail starting with velocity, Vo at closest approach to the Sun we have calculated to be: V@ = [[O.5(l + k) (1.26 x 10^18 Rs^2 - 2.66 x 10^28M]/Mro + Vo^2]^0.5 where ro is the closest approach distance from the center of the Sun. Another useful formula for partially transmitting sails, k in the above equations is: Ra k = -------- (Ra + A) where Ra and A are the fractional sail reflectivity and absorption, respectively. Just when you thought it was safe to put away your physics books... > Think how far one travels at 0.45c during 100 days... > Way further than where the value 1.27E6 Watt/m^2 is valid. > > For the last time, you can't stop a sail that moves with relativistic speeds >by simply using unconcentrated solar light!!! > > (Also if you can stop the sail by using the plain sun, you would also be > able to accelerate it using the plain sun.) Of course you can, but... My idea was to boost quickly to cruise velocity, then take a much longer time to decelerate to make up for the higher cruise velocity. It made sense at first glance, but read on... > > P.S. I can't do any accurate calculations, since the math would soon become > more difficult than anyone would like. A possible solution is to use a > computer program that calculates discrete steps. However it will also show >that it just can't be done. Okay, here is where I eat crow... I did the math, and not only is it possible it is even possible to get up to high relativistic velocities. However, (drum roll and bagpipes playing a dirge) if there are humans on board, the best we can hope for from a purely solar powered ship is only about 0.003c. The initial acceleration required for relativistic velocities would kill people. What is the real bummer here is that if we do boost the sail with additional energy from masers or lasers in order to bring its cruise velocity up to relativistic speeds then there MUST be an additional amount of energy at the other end to decelerate it. Some deceleration is possible, much more than what you had calculated, but nowhere near enough to come down from much over 0.003c. The velocity of the sail decelerated starship a distance rf from the center of the target star is given by: __________________ / 2(J-GMstarMc) Vf= /Vc^2- ------------- \/ rfMc (1) where Vc is the interstellar cruise velocity of the starship, Mstar is the star's mass, and (1 + k) J = ------- (6.3 x 10^17) Rs^2Lf (Rs in meters) 2 (2) In Equation 2, Lf is the ratio of the star's to the Sun's luminosity. This works out to a limit of about 50 AU for any real acceleration/deceleration for a star in the same class as the Sun with similar luminosity. If you assume a mass of 5 X 10^7 for the ship it could be accelerated up to 0.003c and decelerated from 0.003c within 35 AU of each star without ever surpassing 15g. Umm, this is a mission duration of over 1,000 years by the way. I never REALLY liked sails anyway. Lee Parker (o o) ------------------------------------------------------oOO--(_)--OOo--------- I have seen the turnips singing By a lordly cabbage led; I have heard a dewdrop clinging To the rose that bowed her head; I have sniffed at a sonata, I have touched next Friday week; I have tasted a cantata I have smelt a sausage speak. Now of old if I had wildly Made the claims I do today I should soon, to put it mildly, Have been firmly led away; Doctors, acting with decision, Would have taken me in charge; Now they call it television -- And you see, I'm still at large! -- Lucio in the Manchester Guardian

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