rick55 wrote:George Noori had Hoagland on C2C the night of the landing along with a few other guests. I only heard part of the show as I fell to sleep but I recall something about the speed at which the Curiosity landing craft would release its parachute. That was said to be above the speed of sound by one of the guests. I guess that's about 700 mph. That seems a bit fast to consider a parachute, doesn't it?
I posted this question about the parachute and speed of the lander and later, Haiwa posted something relevent but in response to SmokingGun's questions about engine power at the end. They're related but the parachute is an independent component. Heiwa wrote....
Mars atmosphere is pretty thin so parachutes will hardly brake anything coming flying in at 5 000 m/s, so what you need is a rocket engine that can apply an opposite force that stops the mass m (kg) coming dropping down, e.g. if the mass is 1 kg of thing being dropped at 5 000 m/s, the rocket engine must apply 12.5 MJ (megajoule) energy to stop it (and the engine itself).
This dismisses the parachute too quickly since NASA/JPL depended on the parachute as the first braking effort and thus the first one we should look at to discredit the highly renowned CalTech and all of their students, professors, administration, alumni, and related participants in this apparent hoax-- which in itself is astonishing beyond belief. Heiwa's second post about this was very much closer to where we have to go. He wrote
atmospheric pressure on the Mars ground is only 6 hPa compared with a pressure of 1000 hPa on Earth. In spite of this, we are told the parachutes worked
He also says the the initial velocity is 5,000 m/s but I've read on the NASA site somewhere that it was actually 5,900 m/s which is closer to 6,000 m/s if we're going to use round numbers. This is enough of a difference to check. Heiwa goes on to point to the average speed during descent but I don't understand why that's relevent... the deceleration is relevent. Can we brake the momentum of the craft in time to get to where the retrorockets are fired?...which is a separate component of the problem.
Heiwa points to NASA measuring the height of the atmosphere from the center of the planet which is, as he implies, useless and absurd. We want to know the depth of the atmosphere from the solid ground where, presumably, the lander would crash if it didn't slow down sufficiently. He quotes JPL's page where we see it is considered to begin 125 km above the surface. This should be trivial rocket science arithmetic and, in fact, model rocket enthusiasts are able to do the neccessary formula and numbers. The problem I see is that I'm not personally familiar enough with numbers and formulas to verify that they work and neither are most of the public. So the calculation of the parachute drag on the craft will have to be explained to novices as we go along.
The quote, for example, that starts with 5,000 m/s which is actually closer to 6,000 m/s is, in fact, 6 kilometers per second. This is important so the novice can imagine the units as comparable. IF the atmosphere above the ground starts at 125 km, and the craft is travelling at 6 km/s, it's fairly easy for a lay person to see that he has to stop in 125/6 km//km/s or about 21 seconds. That assumes you're dropping straight down, however. Still, it's useful to visualize this much.
Presumably, the 13,000 mph is about 6 km/s
https://encrypted.google.com/search?hl= ... 7brbcpNveE
The "deceleration" would have to include, then, the initial v(i)=6 km/s and the v(f)=0 with elapsed time in seconds as 7X60x= 420s. Here is the formula for acceleration and deceleration....
http://wiki.answers.com/Q/What_is_the_f ... celeration
v(f) - v(i) / t = -6/420 m/s/s = about -0.14 m/s/s
https://encrypted.google.com/search?hl= ... tJ5mUqNQLs
But what does that translate to in terms of something we can intuitively understand as lay novices? I think it was on Noori's show with Hoagland that nite that the deceleration was equivalent to a car going 60 mph stopping in a fraction of a second. Googling the google calculator we get 1 mph = 0.44704 meters per second so 60 mph = 26.8224 meters per second or 60 mph = 0.0268224 kilometers per second.
https://encrypted.google.com/search?q=6 ... per+second
Using the same deceleration formula,
v(f)-v(i) /t = 0 - .0268 m/s / 420 s (the 7 minutes of terror) = -6.38628571 × 10-5 m/s/s.
That figure, -.000063 m/s2 is way off compared to -0.14 m/s2 as a decelertation figure. So Noori's guess was wrong. Instead what if we stopped a car at -0.14 m/s2. What would that be like for a car going 60 mph? I'll do this later if someone else doesn't.
The momentum and weight values looked at by Heiwa earlier are relevent here too but not just yet-- until we visualize the actual velocities involved. I don't want to complicate it too much at a time. We want to embarrass JPL/CalTech students, staff and others publicly and shame them into admitting a hoax. To do that, the explanations here have to be sellable to the public at large in a way that teaches novices and lay people how rocket science works, by the numbers. Any grade 5 level arithmetic book graduate should be able to follow the calculations...any intelligent 12 year old, in other words. The future of the world depends on these kids grasping the hoax by the numbers.
Am I treading the correct path for this audience here or not? We want to discredit JPL right? And we want to do it "logically" so we can brag about it and tell everyone we know... and make it easy for intelligent lay people to follow. We want to shame the professionals and hoaxers out of perpetrating such hoaxes. Obama says he wants American kids to compete with others on the international scence... welll this seems like a good situation to start with. Maybe we could generate a buzz in the nation's grade schools asking high school students to prove that the Curiosity actually could land on Mars. Somehow, the hoax must be exposed to the light of day!