I'm not sure if integrating pressure is the right thing to do.You want to start with DV/DT = Constant x Pressure.

But you've neglected the sun's gravity entirely which is the biggest weakness of the scheme.

Ah yes, 9 days instead of 9 hours.I was integrating pressure because I figured it would just add up over the journey.Note that, if you make the lower limit on the integral 0.01 AU instead of 0.1 AU, you end up with a final velocity of 0.03c. But that's of course assuming the sail doesn't rip apart being that close to the sun.

Mass of sail turns out to be almost negligible if it’s only 1 nm thick:

Thus we could accelerate 100 metric tons to 0.01c just by using a big and thin enough solar sail.

Thanks for the responses. Doesn't hurt to toss some calculations out there!

If we scale up the sail to a side of 1 km, keeping the same thickness

OK so here’s some more calculations:The Ikaros sail, currently deployed, has a thickness of 0.0075 mm, or 7.5x10^-6 m.(http://www.jaxa.jp/projects/sat/ikaros/index_e.html) I couldn’t find the density of the sail, so I just went with Beryllium (1.85 kg/m^3). (On the Ikaros web page they say that they used polyimide, which has a density of 1.43 x 10^3 kg/m^3, but I was pretty sure they coated it with some metal)The area of the Ikaros sail is just under 200 m^2, being 14 m on one side.If we scale up the sail to a side of 1 km, keeping the same thickness, the area is then 10^6 m^2, so the volume of the sail would be:10^6 m^2 * 7.5x10^-6 m = 7.5 m^3And the mass would be:7.5 m^3 * 1.85x10^3 kg = 1.39 x 10^4 kgAssuming the payload is 8.61x10^4 kg, for a total mass of 10^5 kg, we have (based on Jim Davis’ formula for final velocity):Vfinal = sqrt(2 * S * k0 * R0 / m *9.9) = sqrt (2 * 10^6 m^2 * 4.6x10^-6 Pa * 1.5x10^11 m)/10^5 kg * 9.9) = 1.17 x 10^4 m/s, which is nowhere near the speed of light, but it is 2.46 AU/year.

In the longer term.. would a graphene sheets be a good candidate for a solar sail?

you also need to account for the structural mass, which averages about 15% of total mass. Furthermore, you need to figure in a carbon fiber grid as a form of rip-stopping.The sail that calculated reaching 0.03c was something like 500 sq km in area. This is without any sort of laser augmentation. I spoke to the author of that paper about that possibility and he hasn't yet done any calculations involving thrust augmentation from lasers.Note also that in order to decelerate from that speed once your sail reaches its target star system, the star needs to have at least the same level of output as our sun or the sail will overshoot its target with excess velocity.

Quote from: mlorrey on 08/29/2010 01:56 AMyou also need to account for the structural mass, which averages about 15% of total mass. Furthermore, you need to figure in a carbon fiber grid as a form of rip-stopping.The sail that calculated reaching 0.03c was something like 500 sq km in area. This is without any sort of laser augmentation. I spoke to the author of that paper about that possibility and he hasn't yet done any calculations involving thrust augmentation from lasers.Note also that in order to decelerate from that speed once your sail reaches its target star system, the star needs to have at least the same level of output as our sun or the sail will overshoot its target with excess velocity.What journal was this published in? What's the author's name?

Reaching extra-solar-system targets via large post-perihelionlightness-jumping sailcraftGiovanni Vulpetti