**P**= v^3 * CD =

**L^3/2 * CD * CL^-3/2**.

- proportional to mass^1.5,
- proportional to the drag coefficient and
- inversely proportional to the lift coefficient^1.5.

Posted in airplane, Models, Motivation, Navelgazing, Science, Transportation, tagged Aircraft, Conceptual Design, cruise, HALE, Loiter, UAV on Tuesday 2010.03.09| Leave a Comment »

Karoliina has some thoughts on plane design, looking at High Altitude Long Endurance (HALE) UAV:s as inspiration for high L/D craft, to ultimately cruise at fast speed with little power. I disagree somewhat, and I’m sketching out why, below.

Probably the drones, like sailplanes, want low sinkrate and high L/D is secondary, because they need to just stay aloft and not go anywhere.

Power needed is P = v*D (we assume). Since D = CD*v*v, P = v^3*CD.

Lift is L = CL*v*v so v = (L/CL)^(1/2). (Note this CL is different from the L = 0.5*rho*A*cl*v^2, so CL = 0.5*rho*A*cl. It’s more practical here.)

Power thus is **P** = v^3 * CD = **L^3/2 * CD * CL^-3/2**.

The lift equals mass, so the power needed is

- proportional to mass^1.5,
- proportional to the drag coefficient and
- inversely proportional to the lift coefficient^1.5.

This means the lift coefficient CL needs to be large for the craft to be able to loiter for a long time. So long wings and somewhat cambered profiles. A little drag doesn’t hurt as much as low lift so struts are a possibility.

Instead, for a piston cruiser, the L/D needs to be maximized for a certain minimum trip fuel consumption, not per time. Basically, you want to minimize delta_E = P*delta_t = P * delta_x/v so the cost function J = v^2 * CD = L * CD / CL which minimizes at maximum L/D. The CL term is less important compared to the loiterer. As a first guess this should optimize to a less cambered airfoil and smaller or shorter wings. And no struts.

Posted in Architecture, Demotivation, Humor, Motivation, RLV:s, Science, Spacecraft, Suborbital, Transportation, Uncategorized, tagged cruise, hypersonic, hypersonic cruise, rocket on Friday 2009.03.20| 2 Comments »

With some caveats. ðŸ™‚ Let’s assume a rocket is launched, and accelerates to constant speed v_c. Then it stays cruising at this speed and at a constant altitude. Landing is disregarded.

We must modify the rocket equation slightly for the cruise: dm/dt is mass flow, v_ex is effective exhaust velocity, F is the thrust, g is the gravitational acceleration 9.81 m/s^2, m(t) is the mass as function of time, L/D is the lift to drag ratio. If we use the for time, (x is the cruise distance) we can integrate it from start to final mass just like the rocket equation and get the cruise mass ratio: Notice how with increasing cruise speed, the required mass ratio for cruise is lessened. This is because less time is spent in the air and thus the gravity losses are lessened.

But we have to take into account the acceleration to cruise speed as well, which requires some mass ratio as well. We don’t take into account the distance traveled during acceleration, or lift, as the acceleration is a relatively short time and distance phenomenon with rockets that easily optimize to have high T/W.

Now, for the total required mass ratio, we multiply the two mass ratios. Then we search for the minimum total mass ratio by derivating it and searching for the zero point. We get: – **the optimum cruise speed (smallest mass ratio)** Notice how the exhaust velocity cancels out, the optimal speed doesn’t depend on it.

If I calculated right, for a 6000 km transatlantic rocket powered flight with a lift to drag of 7, the best cruise speed for minimum mass ratio is 3 km/s. If you go slower, you waste fuel by hanging in the air, if you go faster, you waste fuel by accelerating too much. I think that’s about Mach 9 at some altitude. This didn’t take into account the deceleration: faster cruise speed takes some advantage there! Even if you shut the engine, it glides further. In real life there are multiple issues:

- acceleration takes time and distance too
- engine T/W size has an effect as well
- there is varying mass during flight .which reduces lift needs with time o which in turn effects L/D as you go higher or reduce AoA .which also requires throttling
- And a million other things.

Also L/D 7 is probably much too good. Oh, and in the transatlantic case, mass ratio required with exhaust velocity 3 km/s would be 7.

To be more complete, Ian Woollard already posted musings on Arocket how this boost-cruise is really quite inefficient from an overviewing energy viewpoint: it would be best to burn the rocket fuel at the fastest possible speed – that means right at the start. Â Then you use the speed and altitude reserve to glide to the goal. To really have a better look at all the problems with this (like really with boost-cruise too), one would need some hypersonic polars of some real vehicle shapes – which cL and L/D at which altitude and speed and AoA. Â The boost-glide could have problems as well, if the velocity is very sharply downwards (as you can’t accelerate fast in the atmosphere for fear of melting) and you need a high cL to turn it around to horizontal. And you also experience high g lateral forces there – at 3 km/s or approximately Mach 9, a 40 km radius turn requires 23 gees of centripetal acceleration. Ouch!

Remember kids, this was just a quick post, nothing serious – don’t try hypersonic rocket cruise at home!

Posted in Models, RLV:s, Uncategorized, tagged cruise, hypersonic, Slug, transatlantic, v prize, virginia, wallops, X-24C on Wednesday 2008.01.23| 22 Comments »

The Virginia Spaceport at Wallops (east coast of USA, south from New York) advocates want to create a prize to foster use for the spaceport, and have floated an idea of a “one hour to Europe” style prize, the V Prize, but the rules aren’t yet finalized.

The requirements are quite tough. About 6000 km in one hour implies a speed of over Mach 6. There’s some discussion here too.

It’s probably a useless exercise but fun nevertheless, so I propose a notional atmospheric (as opposed to purely ballistic ICBM style) vehicle design for the prize.