The 80/20 rule

[wdobner]

Since you guys are working through this on my account I guess I'll offer up that a visit to onlineconversion.com gave me 650,000 lbf being 294,835.04 kg.

 

[barrykirk]

OK, that is 1.34 G's <br /><br />But I subtracted one because the net acceleration is 0.34 G's positive.<br /><br />

 

[barrykirk]

Burn time of that first stage is 249 seconds. So, it goes through that fuel awefully quickly.<br /><br />I reviewed the video footage from the launch and the commentator mentioned that they hit mach 1 at about T+80 seconds.<br /><br />Assume that mach 1 is about what 1,100 Feet per second.<br /><br />For an average acceleration of 13.75 feet per second per second during the first 80 seconds of flight. One G is 32 Feet per second per second, so yes we are doing fractional G's during the first 80 seconds. The accleration will pick up quickly after that, because most of the mass of the vehicle is the first stage fuel and that goes pretty quick.<br /><br />That goes with what I've been saying. They could use a lot less fuel if they went with a higher thrust engine. Even one buring a lower ISP fuel like RP-1. The gravity losses were staggering.<br /><br />Now on upper stages where gravity and drag losses are less. Thrust is less of an issue. Look at the 800+ seconds of burn time with the second stage. That is where LH2 is really useful.

 

[nacnud]

Well tanks are cheap(ish), for a max payload for a given engine just keep adding tankage untill it can only just get off the pad. The Delta IV is really slow to leave the pad compared to the shuttle. It might be a good candidate for an assisted launch.<br /><br />One other think to note is the mess it would make of the launch pad if it had an engine shut down early in flight, it hangs around a loong time.

 

[barrykirk]

While going over those numbers. 1100 feet per second at T+80 seconds. One has to wonder what altitude that occurs at.<br /><br />Assume constant accel which I know is not true. But it might be a close first approxmation. Let's further approximate and say 14 Feet / second /second<br /><br />At t= 80 seconds<br /><br />d = 1/2 a t^2<br /><br />d is about 45,000 feet at t = + 80 seconds.<br /><br />Well, that's good and bad. It means that max q is going to occur at higher altitudes with thinner air and less drag.<br /><br />That is one of the pluses of that slow initial acceleration.<br /><br />I don't think that max q is necesarrily the point of transition between subsonic and transonic, but I've not seen any info on how that is calculated.<br /><br />I suspect it is much more complex than the simple rocket equation.

 

[chris_in_space]

"I don't think that max q is necesarrily the point of transition between subsonic and transonic, but I've not seen any info on how that is calculated. <br /><br />I suspect it is much more complex than the simple rocket equation."<br /><br />Well it depends. If you're willing to do some basic physics and if you're satisfied with just a "first approximation" it's not that complicated.<br />You just take the force giving aerodynamic drag: <br />F=1/2*rho*S*Cx*v^2<br /><br />with <br />-rho the atmospheric air density (in kg/m^3 for example) following the standard atmospheric model depending on z (altitude). The form of rho function of z is a decreasing exponential in the standard atmospheric model (some research on the net and you can find the detailed funtion I think). <br />-S is the area of the cross section of the rocket<br />-Cx the "drag coefficient". Let's say it's the same as that of a bullet (I think it can be approxiamtley found on the net too)<br />-v the velocity of the rocket. You have it from your acceleration equation. The problem is you have to put it funtion of z(the altitude) and not of t (time). So you inject t function of z in the velocity equation<br /><br />After all this you now have F(z) which is the drag funtion of the altitude. You plot your funtion and look where the maximum is.<br /><br />Well I agree maybe it's not so very simple either (my explanations are a bit short...) but with some google you should get it right I hope.<br />The other problem is that it's just a "first appoximation"...<br />

 

[tap_sa]

Is drag coefficient a constant or a function of speed? For instance is there a sudden change in it when changing from subsonic to supersonic? If it changes are there any good thumb rules for subsonic-supersonic-hypersonic transitions?

 

[barrykirk]

So, for any given rocket, S and Cx are constants.<br /><br />What would be nice would be a chart showing acceleration versus Force of Max Q.<br /><br />And another chart showing acceleration versus altitude of Max Q.<br /><br />I know that these charts would only be crude approximations.<br /><br />What I'm getting at is this. If the rocket launches from sea-level, then the higher the acceleration the lower the total loses from gravity would be, but the higher the loses from drag would be.<br /><br />At what acceleration do you get the minimum losses?<br /><br />Now, of course I know that acceleration is not a constant. In fact it increases as the rocket burns off fuel.<br /><br />Actually, now that I'm lookig at it. I don't know what I'm asking.... LOL

 

[propforce]

<i>"...Is drag coefficient a constant or a function of speed? ..."</i><br /><br />Yes. So is the Lift Coefficient. <div class="Discussion_UserSignature"> </div>

 

[chris_in_space]

Drag coefficient is function of speed.<br /><br />Good thumb rules are:<br />-Cx is constant (independent of speed) during all the subsonic part<br />-Cx is constant and slightly higher than the subsonic number during hypersonic part<br />-elsewere its a bit complicated...<br /><br />The figure here after shows a typical Cx graph (function of number of Mach).<br />For a rocket Cx would normally be a bit under the cone shape (let's say substract 0.05 of the cone shape Cx). So you can define Cx for rocket calculations funtion of speed too. The model is getting more and more complicated so I don't now if it's really worth it keeping in mind that big approximations were done for the speed acceleration calculus (x% drag and gravity losses).<br />

 

[barrykirk]

Well, there are many different ways to skin a cat or get to orbit.<br /><br />There are tradeoffs... I was just wondering what those tradeoffs are.<br /><br />As an example. if you had a large enough and powerful enough airplane, the old Saturn V with the whole Apollo shebang could have been airlifted and launched from an airdrop. Assuming it had been designed for an airdrop of course.<br /><br />Since it would have started at a higher altitude, it would have used less fuel to perform it's mission for the following reasons.<br /><br />1) It would have been ever so slightly higher out of the earth's gravity well. <br /><br />2) Rocket engines get more efficient as the air pressure drops. They reach peak efficiency in a vacuum. <br /><br />3) It would have less air drag losses.<br /><br />How much would that have contributed to a weight reduction? A lot of that depends on the drop altitude and speed and I don't have the numbers.<br /><br />I can say that for something that size, the airplane would have been huge.<br /><br />Certainly air drop has been used before for small stuff and will probably been more common in the future. Just how much of a benefit does it provide?