arkie87

Thanks for the reply. I found the actual (non peer-reviewed) article: here I think what's needed is the fact that the author computes that there are 210 civilizations out there also broadcasting for the same amount of time. If you compute (1500/ 32000)^2*210 you get approximately 0.5.

I stumbled upon this article from Cornell, claiming that we can expect to encounter alien life within 1500 years. The basic idea is that once our signals reach half the volume of the Milky Way, it's reasonable to assume that an alien civliation will detect them. In the words of the author(s): " Combining the equations for the Fermi Paradox and the mediocrity principle, the authors suggests Earth might hear from an alien civilization when approximately half of the Milky Way Galaxy has been signaled in about 1,500 years. " Now i'm no expert, but the diameter of the milky way is 100,000 ly. How is it possible that within 1500 years, radio signals will reach half the volume? Am i crazy? Link to Article

Check your trim (alt+X resets). If you look on the bottom left, you should be able to see what position your roll, pitch, and yaw actuators are at. If they arent zeroed with SAS off, then hitting alt+X should help.

I was thinking something similar regarding your contract idea, though not related to "on-rails" drag. Basic idea would be a craft or asteroid that just entered Kerbin's SoI on a collision course. You have to launch a craft and intercept it before it impacts Kerbin.

thanks for the answer!

Exactly. Snark's logic only works if we start above the drag threshold, but on Duna, we are always below it (that's why we come in careening at 800 m/s at 3 km above the surface)

By this, i assume you mean: q = 1/2 rho U^2 instead of q= p(1+(γ-1)/2 M^2)^(γ/(γ-1)) - p Also, in KSP, is Mach number calculated from (γ*R*T)^0.5 i.e. does it depend on temperature and atmospheric composition?

I dont really see your point. At the same altitude above surface, one should be able to deploy parachutes at higher speeds at lower density because drag force at a given speed is lower. That's all that matters, and that is the point of this thread.

Exactly, that's why im confused about @Snark's point. At the same altitude, drag should be less on Duna (unless differences in temperature and atmosphere composition contribute to make density the same).

17 km is pretty high still, and might even be below point of maximum Q. It's possible the way the game is tweaked now, the lowest pressure limit for parachutes might prevent deployment before drag picks up above limit. However, in real life, this setting can be changed/lowered-- there is no physical reason for this lower limit (unless drag is so low it doesnt even pull the parachute out). It shouldnt be about pressure. It should be about air density only.

I guess i was ninja'd " as well as tackling an issue where parachutes would overestimate mach effects in low-density situations. " But you listed density as its own variable, so then you dont care about static pressure and temperature

How does drag depend on static pressure and temperature if density is accounted for separately? dynamic pressure = 1/2 Density*Velocity*Velocity? Your point about Duna vs. Kerbin being a red herring since there are other factors which complicate the comparison is well taken though. Either way, it seems like most people agree this is either a bug or an unrealistic simplification made with the limited time the devs have.

When people say Duna's atmosphere is thinner or the pressure is less, they mean the surface pressure is less or the scale height is smaller. But your implied point about density (which is what causes drag, not pressure) is dependent on molar mass and temperature as well as pressure. Surely, the temperature on Duna is less than on Kerbin, so the surface atmospheric density might not be that different (i'm not sure about the composition of Duna's atmosphere though). That said, i'm not sure why you say "how fast you go is a function of air pressure" when you implied above it is a function of density, not air pressure. AFAIK, speed of sound is only dependent on temperature c = sqrt(gamma*R*T); pressure does not play a role, which is slightly counter intuitive.

I landed safely with retroburn. I think indicator finally turned green at around 200 m/s. Still not sure why this speed is LESS than on Kerbin? Seems to me like it should be larger than on Kerbin if drag/stagnation pressure is to blame.

Is that setting not tweakable?

It was a question but your post didnt answer/address it. Increasing air density WILL increase drag if the parachutes are deployed before drag is significant, which is the assumption that is stated in the question (so thats why i didnt understand your answer).

Dont quite understand why you are saying all this. What i said was that if KSP uses ONLY drag as a means to determine whether the parachute will survive, it will tell you that you can deploy it at 60 km @ 8 km/s since density and resulting drag force is so low. That will subsequently cause problems when you get lower and the parachute catches fire and burns up. Yeah, my vehicle is quite heavy. Its a lander with Bob on board and a full tank of gas. I think ill have to burn retrograde until i can deploy chutes.

It seems to me that his mod only affects parts his mod adds; It doesnt change behavior of stock parts,correct? So my ship currently orbitting duna is screwed, right?

Thanks for the answers. So Mach 1 has nothing to do with it, i presume (and Mach 1 is slower on Duna than on Kerbin because of temperature)? The only thing that will destroy a chute is drag force (assuming temperature isnt a problem)? That means the parachute can be deployed in upper atmosphere safely, but they will break off once air density increases and drag force increases? This will cause problems as from a drag force perspective, parachutes can be deployed in upper atmosphere, so the parachute deploy indicator in staging wont be RED. On the other hand, since external temperature is too hot, perhaps they WILL be marked red and it wont cause problems for noobs... How difficult would it be to write a mod that fixes this? Could module manager set a different critical velocity for each planet? It seems like you are suggesting that KSP doesnt break parachutes if the force exceeds a certain threshold, but rather, just checks the speed and maybe altitude. How difficult would it be to write a mod that breaks chutes if the force they WOULD impart is greater than a threshold value, instead of some other criteria?

Sounds like me when i attempted to land on Duna-- freaking out, Except i crashed and then f9'ed Thanks for the reply. I think i worded it poorly. My problem isnt that the parachutes dont slow me down enough and i crash, but rather, that i cannot deploy them at all. My question is: is that realistic or is it a bug?

Thanks for the reply. Same for me-- in past versions, i could deploy chutes whenever. Do you happen to remember how fast you were going before you could deploy chutes in 1.0.5? To what speed did you burn retrograde before you could deploy chutes?

I tried making a craft to land on duna. I put loads of regular parachutes (did not yet unlock drogue shoots in career mode) and small wings to help me steer and slow down. I've managed to get velocity below 300 m/s (i think as low as 240 m/s) but the parachute color in staging still didnt change from red, so when i deployed, the chutes broke. (1) Is it possible to land on Duna (without retrograde burn) w/o drogue chutes? Or am i experiencing a bug/incorrect physics and i should download a mod (real chutes, for instance)? (2) Does this make sense? The density is so low on Duna, could the atmosphere really break my chute off? (3) Does it have something to do with being below Mach 1? Duna's temperature is much lower than Kerbin, so Mach 1 is much slower. Is that it? What am i doing wrong? To clarify: my problem isnt that the parachutes dont slow me down enough and i crash, but rather, i cant slow down enough to be able to deploy them without them being destroyed instantly.

All, I have posted this same question in a cfd-online forums. But i figured i'd repost here, since the question is pretty basic, and there should be a lot of external flow people here (and developing internal and external flow are pretty similar for really short channels). Anyway, I am trying to simulate developing flow between two parallel plates (2-D simulation), and compute the dimensionless developing (Darcy) friction factor Reynolds product over fully-developed value of 96 (f*Re/96) as a function of dimensionless length down the channel (X^ = X/(Re D)). From the literature, it seems that the curve of f*Re/96 vs X/(Re D) should be constant and independent of Re; however, in my simulations, in multiple softwares (ANSYS Fluent, COMSOL, my own Matlab code), i find that the curves vary with Re, and increasing Re decreases f*Re/96 for the same X/(Re D). I compute f two ways, both of which give nearly identical results: (1) dP/dx = 1/2 f/D rho V^2; solve for f (2) f = 4*mu*du/dy|w / (rho V^2) (factor of 4 to convert from Fanning to Darcy) p is static pressure, but ive tried total pressure as well, and it doesnt fix the problem. Is this correct? What am i doing wrong? Would appreciate any help,

What about latent heat? - - - Updated - - - This wastes water and costs extra money. I am looking for a free solution, that doesnt make more dirty dishes. - - - Updated - - - This does work, but makes two dirty dishes. I'm not convinced the second idea would work though... maybe i need to try it... - - - Updated - - - creates more dirty dishes - - - Updated - - - Wow. And your actual solution misses the point and doesnt resolve the issue. If i wanted to divy up the soup into 100 tupperware containers, i would just do that.

True, perhaps i should have specified: i am looking for passive ways. I am lazy and dont want to have to sit there doing anything