How do you calculate the maximum altitude if you know the speed for a suborbital rocket? You would go straight up and omit all propellant and rocket weight kind of things.

How high could you go with an orbital speed rocket?

Quote from: M129K on 09/29/2013 06:58 PMQuote from: baldusi on 09/29/2013 06:06 PMQuote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.If I understand it correctly, t is total time? If so, thank you! Seems very accurate.Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T the function over dt.

Quote from: baldusi on 09/29/2013 06:06 PMQuote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.If I understand it correctly, t is total time? If so, thank you! Seems very accurate.

Quote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.

I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.

Quote from: baldusi on 09/30/2013 01:05 AMQuote from: M129K on 09/29/2013 06:58 PMQuote from: baldusi on 09/29/2013 06:06 PMQuote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.If I understand it correctly, t is total time? If so, thank you! Seems very accurate.Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T the function over dt.Thanks... I guess I'll have to open my math books again.

Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).g: 9.8*sqrt(R)/sqrt(R+h)R: 6371 (in km, the average diameter of the Earth)

Quote from: baldusi on 10/13/2013 12:24 AMThrottle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).g: 9.8*sqrt(R)/sqrt(R+h)R: 6371 (in km, the average diameter of the Earth)AIUI the symbol "g" is usually used for the standard acceleration due to gravity near the surface (about 9.8 m/s/s), not the acceleration anywhere else. More importantly the acceleration at altitude is (assuming spherical Earth): (9.8 m/s/s)*(R/(R+h))^2 --- i.e. square, not square-root. Gravity is proportional to distance raised to the -2 power (i.e. inverse square law): http://en.wikipedia.org/wiki/Newtonian_gravity .Edit: also that's the radius of Earth, not the diameter.

If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"

I like to round 9.8 to 10. Accelerating vertically for 3 minutes would incur a gravity loss of about 180 seconds * 10 meter/sec^2. -- about 1.8 km/s.

My point and the reason for this question in the first place is: Doesn't physics require a minimum gravity drag of 1.9 km/s delta-V, no matter what rocket or canon you use? It follows that people who bandy about a gravity drag of less than 1.9 km/s for their rocket are wrong or more charitably, are being overly optimistic.Edit add - But I don't understand your example of shooting the shell horizontally. Raising the perigee I got, but getting around the need to add the potential energy of altitude went right over my head.

What is truly counter intuitive is the result of adding 950 J kinetic energy to reap 1900 J of potential energy.

That potential energy would convert to 1.91 km/s if it could respond to the gravity field and fall, (your 1900 J).

Atmosphere does not enter because of course you could repeat the same steps while raising the 200 km orbit to a 400 km orbit with very little difference in the delta-V’s and arrive at a very similar result.

So, back to finding the minimum possible gravity drag for a launch vehicle, does your example above mean that gravity drag must always exceed 0.48 km/s? Even though it is not a very achievable minimum, it is good to know that gravity drag numbers larger than that are subject to reduction by “What,” more powerful boosters and wiser choice of launch trajectory profile. Traded against atmospheric drag of course.

We're attaching different meanings to the term.

What I call gravity drag (or gravity loss) is loss incurred while doing an acceleration with a vertical component.On worlds with an atmosphere, a vertical ascent must be made to get above the atmosphere.

... moon, ... snip

QuoteWe're attaching different meanings to the term.Are you sure? My position was that when the vertical component of displacement was considered, ~1.9 km/s delta V is needed to reach 200 km.

Maybe I should have asked for a minimum gravity drag trajectory.

Alright, I'm having some trouble calculating the ISP of certain rocket engines. As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight.