================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Mon, 3 Jun 1996 11:50:47 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re: morephysics (short)Cc: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUOk, I can't find a copy of the rocket equation set up for relitivistic speeds.Could someone forward a copy to me. Oh, should the exaust velocity be considered in ship time or 'real' time? KellyAt 6:37 PM 6/2/96, Timothy van der Linden wrote: >>I went over the Bussard paper and found 14E6 as the upper exaust vel forp + 11B. Which obviously isn't what I'm getting. Eiather I miscalculatedthe watts per Kg (I sent my equations for that in another letter) or Imisread Bussards paper. (I suppose I could write him.) I'ld hate to writeoff about 15% of spec impulse if I didn't need to. (See table below.) >>If you know how to get the exaust speed from Mev numbers let me know.I'm sure those numbers from the papers are correct. >Watch carefully:>The reaction as found in my tables book: >H + 11B --> 3 4He + 8.7 MeV>Let's assume energy conversion is 100% and ALL reaction products willbe accelerated in the 100% effective lineac. >First turn MeV in to Joules:>8.7 MeV = 8.7E6 * 1.602E-19 = 1.394E-12 Joule >Then determine the weight of the particles in kg: >H + 11B is approx 12 u (u=atomic mass unit) >12 u = 12 * 1.661E-27 = 1.993E-26 Kg>Thus the energy per kg is:>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg>(Don't use much more significant numbers, if you do use them you need todefine some things more accurate and cannot simply think that 11 Hweighs as much as 11B.)>Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder howBussard can get a higher velocity, unless he dumps some of the reactionmass instead of accelerating it.>>I was working up the following table of the fuel mass ratios needed toget to or from certain speeds given the exaust velocities. >>NEW numbers>>Fuel --> Exhaust>>Vexh	75E6m/s	100 E6m/s 125 E6m/s 150 E6m/s>>p + 11B --> 3 4He>>11,800,000 m/s	576.0	4,790.0 39,900 332,000>>6Li + 6Li --> 3 4He>>(Combined)>>17,800,000 m/s	67.6	275.0	1,120	4,570>>3He + 3He --> 4He + 2 p>>20,300,000 M/s	42.5	138.0	472	1,620>>De + 3He --> 4He + p>>26,500,000 m/s	16.9	43.5	112	287>Of course ;) I get completly different numbers (at least for higher Vend) >674 2,680 82,000 1,200,000>73 181 1,728	10,000>44	97	708	3,377>18	34	154	509>I assume this time relativistics is the origin (for 0.5c gamma=1.15 whichmeans its not almost equal to 1 anymore) for our differences >>Amazing how touchy the fuel ratios are to changes in exaustvel/specific impulse. Look at the difference between 6Li + 6Li and 3He +3He! ;) >Well yes, that is because of the energy per kilogram of fuel. Which is 3times higher for 3He + 3He. (This is what my "magic" number f says) >Timothy---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Tue, 04 Jun 1996 00:59:00 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: more physics (short)>>Thus the energy per kg is:>>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg>So we used the same numbers (thou in a different order) and got about thesame number.I guess so.>One suprize, I had that the mass of Protons and neutrons as about 1.673E-27 kg, where you have 1.661E-27. Is that a typo? We came up with thesame velocity, which seems odd if we were using a different constant. I used the definition for 'u', which finds it origin in the mass of a Carbon12 atom. The weight of 12C is defined exactly 12*u. A single proton orneutron is heavier than the protons and neutrons in a atom-core ofmultiple nuclei (hence fusion frees energy).But like I said before, the accuracy of this number only makes sense if youknow the (nearly) exact numbers of atom-masses. >>(Don't use much more significant numbers, if you do use them you needto define some things more accurate and cannot simply think that 11 Hweighs as much as 11B.)>>Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder howBussard can get a higher velocity, unless he dumps some of the reactionmass instead of accelerating it.>No the numbers were not based on an engineering output. That wassupposed to be the direct output from the reactions. Could be I misreadsomething and the 14E6 number was for something else (though I can't seewhat). I don't know, but I've been discussing this with Rex lately and he thinksthat the efficiency of a DIRECT fusion engine may be low. With direct Imean that one doesn't convert the kinetic energy of the reaction productsto electricity.>>I assume this time relativistics is the origin (for 0.5c gamma=1.15which means its not almost equal to 1 anymore) for our differences >??>Why are you geting larger (often much larger) numbers for 75E6m/s, 125E6m/s, and 150 E6m/s, but smaller for 100 E6m/s? Also the deltas don'tseem even. Hum. I suppose I should rerun my numbers with relativisticequations. With our luck I'll come up with a third set of numbers. I think the deltas are exponential, this causes rapid deviations when oneneglects higher terms (ie. use classical instead of relativistic formulas) Ibelieve the relativistic rocket equation is: M=Mo Exp[Vend/(g Vexh)] where g=1/Sqrt[1-Vexh^2/c^2] (Vexh is relative to the ship)>Pity you gave no way to use you magic number. (Like the value for givenfuels?) Also the F number table didn't give the numbers for the fuels,which naturally didn't have even F numbers. Well simply devide c^2 (9E16) by the J/kg numbers (eg.9E16/6.995E13=1287) Compared to 257 for the best fusion fuel we havecome up with. Somewhere in my document there are a few fusion reactions with f, Mevand J/kg. Here is something that looks like it:J/kg	f2H + 3He -> 4He + 1H + 18.3 MeV 3.51E14 257 2H + 3H -> 4He + n + 17.6MeV 3.38E14 267 3He + 3He -> 4He + 2 1H + 12.9 MeV 2.06E14 437 6Li +6Li -> 3 4He	+ 20.0 Mev 1.60E14 564 (Combined)3He + 6Li -> 2 4He + 1H + 16.0 MeV 1.28E14 704 1H + 6Li -> 4He + 3He +4.0 MeV 5.47E13 1645 1H + 11B -> 3 4He	+ 8.7 Mev 6.99E13 1287TimothyP.S. To Kevin, I'm working on your latest letter...================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Tue, 04 Jun 1996 11:26:29 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: CorrectionHi Kelly,After having stared at it for half an hour I typed that relativisitic rocketequation. Now I know why I couldn't believe its simplicity. This morning Istill was thinking about it and suddenly I knew what I had done wrong,well not wrong, but the formula is to no use for you, since it only takesinto account the relativistic exhaust speed and not the relativistic ship(Vend) speed. Since in your calculations Vexh is hardly relativistic theformula is completely useless. To derive the right equation, would take awhile, maybe Rex has it by hand? CU, Timothy================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Tue, 4 Jun 1996 07:59:26 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re: morephysics (short)Cc: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUAt 12:59 AM 6/4/96, Timothy van der Linden wrote: >>>Thus the energy per kg is:>>>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg >>So we used the same numbers (thou in a different order) and got aboutthe same number.>I guess so.>>One suprize, I had that the mass of Protons and neutrons as about 1.673E-27 kg, where you have 1.661E-27. Is that a typo? We came up with thesame velocity, which seems odd if we were using a different constant. >I used the definition for 'u', which finds it origin in the mass of a Carbon12 atom. The weight of 12C is defined exactly 12*u. A single proton orneutron is heavier than the protons and neutrons in a atom-core ofmultiple nuclei (hence fusion frees energy).>But like I said before, the accuracy of this number only makes sense ifyou know the (nearly) exact numbers of atom-masses. Well eiather way I can't remember where I got my atomic mass constantfrom. Eiather one seems to get similar numbers, so its probably not abiggy.>>>(Don't use much more significant numbers, if you do use them you needto define some things more accurate and cannot simply think that 11 Hweighs as much as 11B.)>>>Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder howBussard can get a higher velocity, unless he dumps some of the reactionmass instead of accelerating it.>>No the numbers were not based on an engineering output. That wassupposed to be the direct output from the reactions. Could be I misreadsomething and the 14E6 number was for something else (though I can't seewhat). >I don't know, but I've been discussing this with Rex lately and he thinksthat the efficiency of a DIRECT fusion engine may be low. With direct Imean that one doesn't convert the kinetic energy of the reaction productsto electricity.Why? The velocities seem to be similar to the 'optimum' velocities in yourtable. I thought you figured that by the time you converted the particalsmomentum to elec, and then convert that elec back to the exaust velocityof a mass it would all come out even? (Give or take a lot of tonage ofpower equipment.)Please CC me if you have anything.>>>I assume this time relativistics is the origin (for 0.5c gamma=1.15which means its not almost equal to 1 anymore) for our differences >>??>>Why are you geting larger (often much larger) numbers for 75E6m/s, 125E6m/s, and 150 E6m/s, but smaller for 100 E6m/s? Also the deltas don'tseem even. Hum. I suppose I should rerun my numbers with relativisticequations. With our luck I'll come up with a third set of numbers. >I think the deltas are exponential, this causes rapid deviations when oneneglects higher terms (ie. use classical instead of relativistic formulas) Ibelieve the relativistic rocket equation is: >M=Mo Exp[Vend/(g Vexh)] where g=1/Sqrt[1-Vexh^2/c^2] >(Vexh is relative to the ship)Thanks for the equation. As to my question about why are you geting largernumbers for 75E6m/s, 125 E6m/s, and 150 E6m/s, but smaller for 100E6m/s? My guess would be you calculated for .3c not 1/3rd c again. >>Pity you gave no way to use you magic number. (Like the value for givenfuels?) Also the F number table didn't give the numbers for the fuels,which naturally didn't have even F numbers. >Well simply devide c^2 (9E16) by the J/kg numbers (eg.9E16/6.995E13=1287) Compared to 257 for the best fusion fuel we havecome up with. Best is of course a relative term, and so obviously is the term simple. ;) >Somewhere in my document there are a few fusion reactions with f, Mevand J/kg. >Here is something that looks like it:>J/kg	f>2H + 3He -> 4He + 1H + 18.3 MeV 3.51E14 257 2H + 3H -> 4He + n + 17.6MeV 3.38E14 267 3He + 3He -> 4He + 2 1H + 12.9 MeV 2.06E14 437 6Li +6Li -> 3 4He	+ 20.0 Mev 1.60E14 564 (Combined)>3He + 6Li -> 2 4He + 1H + 16.0 MeV 1.28E14 704 1H + 6Li -> 4He + 3He +4.0 MeV 5.47E13 1645 1H + 11B -> 3 4He	+ 8.7 Mev 6.99E13 1287Thank you. Thats a table that would have made your other table useful. Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Tue, 4 Jun 1996 08:06:32 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re:CorrectionCc: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUAt 11:26 AM 6/4/96, Timothy van der Linden wrote: >Hi Kelly,>After having stared at it for half an hour I typed that relativisitic rocketequation. Now I know why I couldn't believe its simplicity. This morning Istill was thinking about it and suddenly I knew what I had done wrong,well not wrong, but the formula is to no use for you, since it only takesinto account the relativistic exhaust speed and not the relativistic ship(Vend) speed. Since in your calculations Vexh is hardly relativistic theformula is completely useless. To derive the right equation, would take awhile, maybe Rex has it by hand? >CU, TimothyAh. Good point. Thanks for the follow up. Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Tue, 04 Jun 1996 21:04:41 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: more physics (short)>>>No the numbers were not based on an engineering output. That wassupposed to be the direct output from the reactions. Could be I misreadsomething and the 14E6 number was for something else (though I can't seewhat). >>I don't know, but I've been discussing this with Rex lately and he thinksthat the efficiency of a DIRECT fusion engine may be low. With direct Imean that one doesn't convert the kinetic energy of the reaction productsto electricity.>Why? The velocities seem to be similar to the 'optimum' velocities inyour table. I thought you figured that by the time you converted theparticals momentum to elec, and then convert that elec back to the exaustvelocity of a mass it would all come out even? (Give or take a lot oftonage of power equipment.)>Please CC me if you have anything.It has to do with the fact that one needs to build up pressure in the engineso that all reaction products get a more or less even velocity. Then theparticles can be directed to a "hole" (the outlet) in the engine. It is thequestion if the pressure can be build up, since a lot of the energy will beradiated away, we may be talking about million degrees here. Maybeparticles can be contained but black body radiation is hard to keep in. Thisway there will probably be to much losses, therefore it may be best tocapture that heat right away and turn it into electricity. Timothy================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Tue, 04 Jun 1996 21:04:32 +0100To: kgstar@most.fw.hac.comFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Java calculations(I'd sent this letter only to your home address, a few hours ago) Hi Kelly,Today I've been busy writing a JAVA-script to do the calculations that arein my document about "relativistic spaceflight calculations". I'm not sure if you have Netscape 2.xx, because that is what you need tomake the thing work.I include here the HTML-code with the JAVA-script, please try it out. Notethat it is a rough setup, nothing fancy (not even user friendly). If it is succesful, I will soon make a clearer version and put it on the web. TimothyTry pasting the following into a text-file (ah well, you know...)============================================================================= <HTML><HEAD><TITLE>Relativistic spaceflight calculations</TITLE> <SCRIPT><!-- Begin to hide script contents from old browsers. var newline=nullif (navigator.appVersion.lastIndexOf('Win') != -1) { 	newline="\r\n"}else {	if (navigator.appVersion.lastIndexOf('Mac') != -1) {		newline="\r"}	else {		newline="\n"}}function ff(Vexh,Vend) {	//Formula that needs to be solved}	with(Math){	VVexh=Vexh/c;	VVend=Vend/c;	cc=1; //Set speed of light to 1	Temp1=2*pow(VVexh,2)*VVexh;	Temp2=pow((cc+VVend)/(cc-VVend),cc/(2*VVexh));	Temp3=cc*sqrt(pow(cc,2)-pow(VVexh,2))*(sqrt(pow(cc,2)-pow(VVexh,2))-1); 	Temp4=log((cc+VVend)/(cc-VVend));	}	return(Temp1-Temp2*(Temp3*Temp4+Temp1)); }function Sign(x) {	// Return the sign of x	if (x<0) {signvalue=-1}		 else {if (x>0) {signvalue=1}		 		else {signvalue=0}}	return(signvalue);}function BestVexh(Vend) {	//Numerical solve using the bisection method} 	Start =0;	Finish=c;	Dif=10000;	while (Dif>1000) { // Accuracy=1000	 Mid=(Start+Finish)/2;	 Dif=Finish-Mid;	 if (Sign(ff(Finish,Vend))*Sign(ff(Mid,Vend))<=0) 	 {Start=Mid} else{Finish=Mid}	 }	return(Mid);}function compute(obj) {	with (Math) {	c=299792458; //Speed of Light	f=eval(obj.f.value);	Va=eval(obj.Va.value);	Vb=eval(obj.Vb.value);	Mo=eval(obj.Mo.value);	a=eval(obj.a.value);	Vend=Va*c;	Vexh=Vb*c;	g=1/sqrt(1-pow(Vexh/c,2));	x=Vend/c;	AT=log((1+x)/sqrt(1-pow(x,2))); // ArcTanh[x] 	TT=c/a*AT	t=0;	if ((f-1)*(g-1)<1)	 {// Formula (8.1) (12.1) (14.1)M=Mo*exp(a*(TT-t)/(Vexh));Ek=Mo*(g-1)/g*(exp(a*TT/Vexh)-1)*pow(c,2);t=0;P1=pow(c,2)*Mo*a/Vexh*(g-1)/g*exp(a*(TT-t)/Vexh);t=TT;P2=pow(c,2)*Mo*a/Vexh*(g-1)/g*exp(a*(TT-t)/Vexh); Method=1;}	else	 {// Formula (8.2) (12.2) (14.2)M=Mo*exp((f*g-f)*a*(TT-t)/(g*Vexh));Ek=Mo/f*(exp(a*TT/Vexh*(f*g-f)/g)-1)*pow(c,2);t=0;P1=pow(c,2)*Mo*a/Vexh*(g-1)/g*exp(a*(TT-t)/Vexh*(f*g-f)/g);t=TT;P2=pow(c,2)*Mo*a/Vexh*(g-1)/g*exp(a*(TT-t)/Vexh*(f*g-f)/g);Method=2;	}	// Calculate several optimum parametres 	BestF=1+1/(g-1);	if (f==1){BestG=99999} else {BestG=1+1/(f-1)}	BestV1=c*sqrt(pow(BestG,2)-1)/BestG;	BestV2=BestVexh(Vend);	obj.Method.value = Method;	obj.M.value	= M;	obj.Ek.value	= Ek;	obj.P1.value	= P1;	obj.P2.value	= P2;	obj.BestF.value = BestF;	obj.BestV1.value = BestV1/c;	obj.BestV2.value = BestV2/c;	// Optimal posibility	if (BestV1<BestV2) {BestV=BestV1} else {BestV=BestV2}	obj.BestV.value = BestV/c;	g=1/sqrt(1-pow(BestV/c,2));	BestF2=1+1/(g-1);	if (BestF2>f) {obj.BestF2.value = "For this Vend you can just as welluse f="+BestF2;} else {obj.BestF2.value=""}	}}// End the hiding here. --></SCRIPT></HEAD><BODY><FORM NAME="RSC">Fill out this form<P>	f=<INPUT TYPE="text" NAME="f" VALUE="257" SIZE=4	onChange="compute(this.form)"></input><BR> 	Vend=<INPUTTYPE="text" NAME="Va" VALUE="0.3" SIZE=4	onChange="compute(this.form)"></input>c<BR> 	Vexh=<INPUTTYPE="text" NAME="Vb" VALUE="0.2" SIZE=4	onChange="compute(this.form)"></input>c<BR> 	Mass of thestarship=<INPUT TYPE="text" NAME="Mo" VALUE="1" SIZE=4	onChange="compute(this.form)"></input>kg<BR>	acceleration=<INPUT TYPE="text" NAME="a" VALUE="9.8" SIZE=4	onChange="compute(this.form)"></input>m/s<BR> 	<p>	<INPUT TYPE=button SIZE=8 NAME="recalc" VALUE="Recalc"onClick="compute(this.form)"></input>	<p>	<TABLE BORDER>	<TR><TD>Method</TD><TD></TD><TD><INPUT TYPE=text NAME="Method"	SIZE=10></input></TD></TR>	<TR><TD>Initial mass (kg)</TD><TD>M</TD> 	<TD><INPUT TYPE=textNAME="M" SIZE=10></input></TD></TR> 	<TR><TD>Total energy needed (Joule)</TD><TD>Ek</TD> 	<TD><INPUTTYPE=text NAME="Ek" SIZE=10></input></TD></TR> 	<TR><TD>Initial Power (Watt)</TD><TD>P[0]</TD> 	<TD><INPUTTYPE=text NAME="P1" SIZE=10></input></TD></TR> 	<TR><TD>Final Power (Watt)</TD><TD>P[T]</TD> 	<TD><INPUTTYPE=text NAME="P2" SIZE=10></input></TD></TR> 	</TABLE><HR>Intermediate results (for experienced use)<BR> Best f for current Vexh, f =<INPUT TYPE=text NAME="BestF" SIZE=10></input><BR> Best Vexh for current f, Vexh = <INPUT TYPE=text NAME="BestV1"SIZE=10></input>c<BR>Best Vexh for current Vend, Vexh = <INPUT TYPE=text NAME="BestV2"SIZE=10></input>c<BR>The lowest Vexh is the optimum!<BR><HR>Optimal posibility<BR>In the next try for the same Vend, use Vexh = <INPUT TYPE=textNAME="BestV" SIZE=10></input>c<BR><INPUT TYPE=text NAME="BestF2" SIZE=55></input><BR> </FORM></BODY></HTML>================================================================X-Sender: S9421793@mail.student.utwente.nl (Unverified) Mime-Version:1.0Date: Wed, 05 Jun 1996 18:18:22 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: USING A ELECTROMAGNETIC BEAM TO PROPELL A STARSHIP Kevin wrote:>>I tried to keep it in the middle (laser or maser), at some places I evenused EM-beams. I don't understand the explanation with your numbersabove: You say the cost of a beaming station increases because theconversion from sunlight into electricity is ~ 10%, but masers don't needelectricity too? >What I mean is that while there could be a way to turn sunlight to laserlight without going through an in-between step like electricity, there isno way to do this for masers. The sun's microwave output is just too dimcompared to its visible output.Huh? I got the idea that you preferred maser because it is so efficient,now your saying that laser may not need an in-between-step, from which Iwould conclude that laser is a preferred method. Oh wait, you are pointing at two things here: maser: needs an extra step,easy to convert to electricity laser: may not need an extra step, less easyto convert to electricity Did I get it right?>>I can understand that phasing is easier with longer wavelengths, so Iwill >it's not just the longer wavelengths, a lot of this technology is availabletoday in the form of radar installations and radio telescopes. I see, do you know if it is difficult to transfer this technology to shorterwavelengths?If it is not too difficult it would be worth trying, especially because thetechnology that is available now will need a lot of redesigning (if notcompletely new ideas) anyway.>>>>1 The beaming station very likely needs to be build on amoving/rotating object like a planet, moon or asteroid.>>>possible solutions: 1) Multiply the number of separate beaming stations >>I'm not sure about (1), because a beaming station already consists ofmany smaller masers. Isn't this the same as increasing the size of thebeam? >Not really. consider the following 16 element maser array Ah, now I see, I already assumed every single maser had his own trackingand aiming device. I'm not sure however how independent they would be,but I assume one could build in a rule that not all maser-unit are allowedto move at the same time.>All I know is that by changing the phase of the separate elements (all ofthe Number 4s above, one can change the direction that the EM rad goes offin. I would think that phase adjustments would be more difficult foroptical systems than for microwave. The elements phases are not allchanged the same way at the same time, so yes, I think it is aninterference effect, both constructive and destructive.But what has the upper hand? The constructive or destructive side... (Thiswas the reason for my inquiry.)>>I think I was seeing a phased array as a problem on itself here. >I think pahes array comes with a different set of problems, like not beingable to completely cancel Em going off in some unwanted direction. Thisends up costing power but the advantages of finer control and reduction ofself interference effects should more than make up for a slight powerdrain I guess here you answered my question just above... >>>Solution: Phased array. By conecting two transmitters many hundreds ofkilometers apart, one can simulate a single arperature with the samebaseline.>>I cannot believe this is completely true. More (phased) sources canindeed decrease diffraction effects, but does one extra source have somuch effect? Does anyone know this for certain by head? Otherwise I'dneed to calculate it myself (which I don't like at the moment). >Again, we already do this for space craft. the Magellan probe to Venus forexample. many of the radar pictures were made using synthetic arperature.Instead of multiple antennas, different images from closely spaced timesalong the same orbit were used.I already thought this was the base of your misunderstanding. For what Iknow (I'm not 100% certain) this method only increases resolution (in thesense of larger angle deviations), but it does not decrease diffraction inthe same way.The term interferometry points at the comparing of two parts of awavefront. The difference in phase is then used in calculations thatincrease the resolution. I know it is a bit fague, but I don't have goodliterature about it at the moment. (Feel free to attack me if you still feelyou are right). >maybe a better explanation... the probe sends out a long pulse of radar theprobe picks up a reflection from the surface, and, as it moves in it's orbit,the signal changes. Computers combine the multiple reflections into onethat looks like it came from an antenna as big as the orbital "footprint" >radio telescopes on opposite sides of the globe routinely are connected to each>other and the result is an antenna as wide as the earth. Called Very LongBaseline Interferometry>>>>9 Red shift is especially important when the starship reachesrelativistic >Okay, so it is an increased cost, and not a technical difficulty It may become a technical difficulty if the shift becomes too large, thenthe reflectivity (and absorption) may change for the worse. While it maynot matter much that some energy is lost, it may matter that a small part(even <0.001%) of high intensity radiation enters the crew space filledwith sensitive computers etc.>>12.>>Yes, I know, but that doesn't decrease the weight of the sail, which maybecome a crucial point in the whole design. I'll add this in the solutionsection.>Agreed. I like Kelly's idea of using 6Li for the sail, very elegant. Unless we need a sail for the way back home... >>14.>>Yes, but where do you put the sending/receiving antenna? If they areinside the beam they are likely to be ionized.>I think a visible laser tube (say CO2 for example) which would be mostlyglass or other non-ionizable material, would be immune to this effect. Ifit becomes a real problem, you can shoot the visible light laser throughthe ~ 1cm holes of a Microwave shield.That would be a solution and a reason to use maser ;) >>Advantages:>>>3) Ship can accelerate continously, taking advantage of the timedialation effect. and providing the crew with a near normal gravityenvironment >>Are other designs like fusion engines not able to do this? (in theory) Theadvantages of time dilations are not clear to me (see also 9) >I think the problem with the fusion rockets, is that in order to acceleratecontinuously, the need planatary sized fuel tanks. Kelly's top speed isabout .4C and that's taking advantage of every trick in the book. I agree, but it is only partly true, since you also need to decelerate usingonboard reaction mass (not necessary fuel). If you accelerate too much,that amount of reaction mass needs to get bigger in order to be able tostop the starship.>The main advantages of>Kelly's hybrid fusion/maser design are>1) lowered cost>2) decell stage is independent of earth. >The advantage of time dialations, is that _all_ stars (within reason of course)>tend to be nearly the same time away (as seen by the crew) and muchsavings of food and other supplies can be acheived. Also, regardless of theactual (earth->based) time of flight, earth (or the return masers) only need to send out about>a two-year long pulse of energy to sustain the flight Oops, last time I told you that the Doppler effect was cancelled for thepeople in the starship, I think I was mistaken there. Note the followingexpression for the Doppler shift, after the arrow I've rewritten it so thatyou can see that there is more than f/gamma in it. 2Sqrt[1 - b ]	f	1f'= f -------------- (where b=v/c) --> f' = ------- ------- 1 + b	gamma 1 + bThat extra term makes it so that the time-dilation cannot make up theenergy drop.So to make up that drop Earth would need to increase the intensity of thebeam. Steve, Rex, am I right here?Timothy================================================================X-Sender: S9421793@mail.student.utwente.nl (Unverified) Mime-Version:1.0Date: Wed, 05 Jun 1996 18:19:10 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: Java calculationsKelly wrote:>Haven't installed it yet, but I'll get Netscape 2.x and try your jave asap. Thanks already.I haven't done any Java programming before, so I don't know all the ins &outs. Tim================================================================X-Sender: S9421793@mail.student.utwente.nl (Unverified) Mime-Version:1.0Date: Wed, 05 Jun 1996 18:19:42 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: New loadKelly wrote:>Good news/ bad news>I uploaded new drafts of all my web stuff into Daves server / W/S. >Bad news, it doesn't seem to work right. At least not all the time. >Anyway. The big changes are some text clean-ups and a set of graphicaltable of content maps. So if you enter>http://165.254.130.92/LIT/InterStellar/project/ Will check it out one of these days (did I say that too someone elsealready?) I uploaded my stuff also, but haven't been able to make it accessable, Ibelieve David is working on it.Tim================================================================X-Sender: S9421793@mail.student.utwente.nl (Unverified) Mime-Version:1.0Date: Wed, 05 Jun 1996 18:19:51 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Engine (EM radiation) problemsKelly wrote:>I don't follow why the presure or hole? I was figuring on the reactorbeing in a parabolic charged bowel that would reflect the particals in thegeneral direction. (About a 20 degree cone should be good.) Otherwisetrying to contain that much power would be hard and HOT. Yes, I believe this pressure thing was something I imagined, and later itseemed that the Bussard engine did not work that way. The main reasonfor that pressure thing was to even out the velocities, this way we wouldnot get very fast and very slow particles all together. Particles with highvelocity have a worse momentum:energy ratio (a new term?). This meansthat if you have some energy and want to make the most velocity(momentum) from it, you get the most of it if you use low exhaustvelocities. Unfortunately this also means that more mass is needed, whichis not preferrable.But assuming a specific amount of mass and a specific amount of energy,you get the most momentum if you give all the mass the same velocity.Having said all this, I haven't the faintest idea how much the efficiency ofengine would rise if one did this in reality. One thing that I'm worried about is the question of how much EM-radiationwould be formed during the reaction, I know you said none, but anycollision of particles (or acceleration, or deceleration of chargedparticles) would create EM-radiation. Since it will not be easy to deflectthis radiation, we need a solution for it.Timothy================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Wed, 5 Jun 1996 12:22:10 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re:USING A ELECTROMAGNETIC BEAM TO PROPELL A STARSHIP Cc:KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUAt 6:18 PM 6/5/96, Timothy van der Linden wrote: >Kevin wrote:>>>I tried to keep it in the middle (laser or maser), at some places I evenused EM-beams. I don't understand the explanation with your numbersabove: You say the cost of a beaming station increases because theconversion from sunlight into electricity is ~ 10%, but masers don't needelectricity too? >>What I mean is that while there could be a way to turn sunlight to laserlight without going through an in-between step like electricity, there isno way to do this for masers. The sun's microwave output is just too dimcompared to its>>visible output.>Huh? I got the idea that you preferred maser because it is so efficient,now your saying that laser may not need an in-between-step, from which Iwould conclude that laser is a preferred method. >Oh wait, you are pointing at two things here: maser: needs an extra step,easy to convert to electricity laser: may not need an extra step, less easyto convert to electricity Did I get it right?The sail system for a microwave beam sail is simpler and lighter than fora laser sail.Of course if your using my hybrid fuel/sail configuration you want ahavyer sail... ;)>>>>>9 Red shift is especially important when the starship reachesrelativistic >>Okay, so it is an increased cost, and not a technical difficulty >It may become a technical difficulty if the shift becomes too large, thenthe reflectivity (and absorption) may change for the worse. While it maynot matter much that some energy is lost, it may matter that a small part(even <0.001%) of high intensity radiation enters the crew space filledwith sensitive computers etc.Given the power levels were tossing about a .001% absorbtion would meltthe ship.>>>12.>>>Yes, I know, but that doesn't decrease the weight of the sail, which maybecome a crucial point in the whole design. I'll add this in the solutionsection.>>Agreed. I like Kelly's idea of using 6Li for the sail, very elegant. Thank you.>Unless we need a sail for the way back home... You make a new, far smaller & lighter, sail before you leave for home. >>>Advantages:>>>>3) Ship can accelerate continously, taking advantage of the timedialation effect. and providing the crew with a near normal gravityenvironment >>>Are other designs like fusion engines not able to do this? (in theory)The advantages of time dilations are not clear to me (see also 9) >>I think the problem with the fusion rockets, is that in order toaccelerate continuously, the need planatary sized fuel tanks. Kelly's topspeed is about .4C and that's taking advantage of every trick in the book. <..and if anyone can suggest other books...> >I agree, but it is only partly true, since you also need to decelerate usingonboard reaction mass (not necessary fuel). If you accelerate too much,that amount of reaction mass needs to get bigger in order to be able tostop the starship.>>The main advantages of>>Kelly's hybrid fusion/maser design are 1) lowered cost>>2) decell stage is independent of earth. I'ld add in technical fesability. We still haven't fiqured out how to brake apure sail ship into a unprepared starsystem. Also power requirements should be far less. >>The advantage of time dialations, is that _all_ stars (within reason of >course)>>tend to be nearly the same time away (as seen by the crew) and muchsavings of>>food and other supplies can be acheived. Also, regardless of the actual >(earth->>based) time of flight, earth (or the return masers) only need to send out >about>>a two-year long pulse of energy to sustain the flight Did you mean to say return masers? Obviously the 2 year pulse bit worksoutgoing, but not incoming, ships.Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Wed, 5 Jun 1996 12:27:21 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re:Engine (EM radiation) problems Cc: KellySt@aol.com,kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUAt 6:19 PM 6/5/96, Timothy van der Linden wrote: >Kelly wrote:>>I don't follow why the presure or hole? I was figuring on the reactorbeing in a parabolic charged bowel that would reflect the particals in thegeneral direction. (About a 20 degree cone should be good.) Otherwisetrying to contain that much power would be hard and HOT. >Yes, I believe this pressure thing was something I imagined, and later itseemed that the Bussard engine did not work that way. The main reasonfor that pressure thing was to even out the velocities, this way we wouldnot get very fast and very slow particles all together. Particles with highvelocity have a worse momentum:energy ratio (a new term?). This meansthat if you have some energy and want to make the most velocity(momentum) from it, you get the most of it if you use low exhaustvelocities. Unfortunately this also means that more mass is needed, whichis not preferrable.>But assuming a specific amount of mass and a specific amount of energy,you get the most momentum if you give all the mass the same velocity.Having said all this, I haven't the faintest idea how much the efficiency ofengine would rise if one did this in reality. >One thing that I'm worried about is the question of how much EM-radiation would be formed during the reaction, I know you said none, butany collision of particles (or acceleration, or deceleration of chargedparticles) would create EM-radiation. Since it will not be easy to deflectthis radiation, we need a solution for it.>TimothyI worry about EM and secoundary reactions in the plasma. We are after alltalking about hellish amounts of power. Even small percentages couldincinerate the engine and ship. (Even a mile of Lithium might not protectthe crew.)Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------==================================================================From: David Levine <David@InterWorld.com> To: "'kgstar@most.fw.hac.com'"<kgstar@most.fw.hac.com> Subject: RE: htm vs htmlDate: Wed, 5 Jun 1996 14:07:40 -0400Mime-Version: 1.0Yes, they need to be default.htmDid you get my previous message (yesterday) about me needing to figureout what's going on with the IP address? -David>---------->From: 	kgstar@most.fw.hac.com[SMTP:kgstar@most.fw.hac.com] Sent:	Wednesday, June 05, 1996 1:58 PM>To: 	David Levine>Subject: 	htm vs html>Is it important that I name my default files default.htm as apposed todefault.html? I noticed you used the former, and I was having someproblems (sometimes) with my default. files. If this sufix (.html vs .htm)is important for some reason I'll change mine. >Kelly>---------------------------------------------------------------------- >Kelly Starks	Internet: kgstar@most.fw.hac.com>Sr. Systems Engineer>Magnavox Electronic Systems Company>(Magnavox URL: http://www.fw.hac.com/external.html) >---------------------------------------------------------------------- Date: Wed, 5 Jun 1996 23:45:30 -0400From: KellySt@aol.comTo: kgstar@most.fw.hac.comSubject: Fwd: LIT web site================================================================Date: Wed, 5 Jun 1996 22:16:04 -0700From: Steve VanDevender <stevev@efn.org> To:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Cc:KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUSubject: Engine (EM radiation) problemsTimothy van der Linden writes:>Yes, I believe this pressure thing was something I imagined, and later itseemed that the Bussard engine did not work that way. The main reasonfor that pressure thing was to even out the velocities, this way we wouldnot get very fast and very slow particles all together. Particles with highvelocity have a worse momentum:energy ratio (a new term?). This meansthat if you have some energy and want to make the most velocity(momentum) from it, you get the most of it if you use low exhaustvelocities. Unfortunately this also means that more mass is needed, whichis not preferrable.This statement bugs me because it is completely contradictory tosomething I worked out a while ago and that you seemed to agree with,which is that higher exhaust velocities are best, and the ideal case isturning all your fuel into zero-mass photons moving at the speed of light.You do want to get all your exhaust products moving backwards, whichimplies confining the fuel reaction and reflecting any forward-movingproducts backwards. You're absolutely going to lose energy to heat if youtry to equalize the exhaust velocity. In any case a gas at a particulartemperature doesn't have a uniform set of particle velocities; it's still astatistical distribution about a mean that is characteristic of the gastemperature.================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Thu, 06 Jun 1996 10:30:55 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: Java calculations>Oh bother,>I thought I'ld have a free copy of Netscape 2 yesterday, since it was afree addition to something I ordered that was delivered yesterday.Actually Netscape didn't come with the product, it comes when you send inthe coupon that came with the product. So it might be a while before I cantry your Java script.You can download netscape from the Netscape site (I assume it there alsofor the Mac):http://home.netscape.com/comprod/mirror/index.html If all goes well, this link should put you right in front of the generaldownload page. All you need to do is select some things from a menu. Then look for Netscape Navigator 2.02I hope you can do this at your work otherwise it may become a problem(2.5 Meg). >How does Java work? Wouldn't the animation hog up bandwith? Java-applets as they are called may eat bandwith, it depends on the natureof the program/applet. For example my Java-thing does not use any otherbandwith than to download the html-page it is put in. Once downloaded ityou can go off-line and use it as much as you like. There may be otherapplications that extract extra information from a remote server (eg.stock-quotes) all the time, those would use some more bandwith. Butgenerally it saves bandwith, because you get (a part of) the program rightat your computer (and thus CPU). This essence has scared some who areafraid that viruses will crawl with the Java-applet inside your computer.At the moment there is a very very small chance this may happen, so don'tworry. Java is written in way that it is almost impossible to do anyserious harm to the computer. Of course some bugs and child diseases needto be removed so that the system becomes watertight. I'm not sure what animations you mean, since a lot of the applets will justbe programmes (sometimes these programmes run in a seperate Window). There is another thing that is around for a while already, it is called VRML(Virtual reality meta link). With this language one can create 3D graphics,the user (downloader of the 3D world) can move around (in real time) inthat 3D landscape and click things (which work just as links these days).I've seen a small demo, and it looks great. I believe the next version ofNetscape (called Atlas) will support this. Well time will show it...Timothy================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Thu, 06 Jun 1996 23:15:25 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: USING A ELECTROMAGNETIC BEAM TO PROPELL A STARSHIP Kevin wrote:>>But what has the upper hand? The constructive or destructive side...(This was the reason for my inquiry.)>Both. Constructive in the direction of TC, destructive in other directions Maybe I'll make a simulation about that some time. >>I already thought this was the base of your misunderstanding. For what Iknow (I'm not 100% certain) this method only increases resolution (in thesense of larger angle deviations), but it does not decrease diffraction inthe same way.>>The term interferometry points at the comparing of two parts of awavefront. The difference in phase is then used in calculations thatincrease the resolution. I know it is a bit fague, but I don't have goodliterature about it at the moment. (Feel free to attack me if you still feelyou are right). >ME? Attack? Never! *8^S>However, If it can produce better resolution, does this help the "jitter"problem?Don't think so, however, those phased arrays (as I think they work) areprobably doing the thing you want.One can focus radiowaves (using interference), by placing antennas in aline at specific distances. Note however that these single antennas areradiating in all directions (in contrary to a laser). I'll check it out in literature if I get a chance. >>Oops, last time I told you that the Doppler effect was cancelled for thepeople in the starship, I think I was mistaken there. Note the followingexpression for the Doppler shift, after the arrow I've rewritten it so thatyou can see that there is more than f/gamma in it. >>2>>Sqrt[1 - b ]	f	1>>f'= f -------------- (where b=v/c) --> f' = ------- ------- >>1 + b	gamma 1 + b>Hmm... so earth would need to roughly double output near the middle ofthe pulse?>where b=.99I guess so. I already thought a while about this (that's the reason for thedelay of the last letter). I'd feel more confident if someone would confirmthis.Timothy================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Thu, 06 Jun 1996 23:15:32 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: USING A ELECTROMAGNETIC BEAM TO PROPELL A STARSHIP Reply to Kelly>>Oh wait, you are pointing at two things here: maser: needs an extra step,easy to convert to electricity laser: may not need an extra step, less easyto convert to electricity Did I get it right?>The sail system for a microwave beam sail is simpler and lighter thanfor a laser sail.Yes, that would be a pro.>Of course if your using my hybrid fuel/sail configuration you want ahavyer sail... ;)Laser forth and maser back?>>It may become a technical difficulty if the shift becomes too large, thenthe reflectivity (and absorption) may change for the worse. While it maynot matter much that some energy is lost, it may matter that a small part(even <0.001%) of high intensity radiation enters the crew space filledwith sensitive computers etc.>Given the power levels were tossing about a .001% absorbtion would meltthe ship.That's exactly what I mean. It is very well possible to make somethingalmost 100% reflective for one wavelength, but the better for one, theworse for different wavelengths.>Did you mean to say return masers? Obviously the 2 year pulse bit worksoutgoing, but not incoming, ships.Incoming ships need less energy than outgoing ships (because ofblueshift), does it really matter how long the beam is on? Timothy============================================================================= Subject: Re: Engine (EM radiation) problems Reply to Kelly>>One thing that I'm worried about is the question of how much EM-radiation would be formed during the reaction, I know you said none, butany collision of particles (or acceleration, or deceleration of chargedparticles) would create EM-radiation. Since it will not be easy to deflectthis radiation, we need a solution for it.>I worry about EM and secoundary reactions in the plasma. We are after alltalking about hellish amounts of power. Even small percentages couldincinerate the engine and ship. (Even a mile of Lithium might not protectthe crew.)OK, then we're back on one line...Kelly================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Thu, 06 Jun 1996 23:15:18 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Energy:momentum ratioTo Steve,>Timothy van der Linden writes:>>Yes, I believe this pressure thing was something I imagined, and later itseemed that the Bussard engine did not work that way. The main reasonfor that pressure thing was to even out the velocities, this way we wouldnot get very fast and very slow particles all together. Particles with highvelocity have a worse momentum:energy ratio (a new term?). This meansthat if you have some energy and want to make the most velocity(momentum) from it, you get the most of it if you use low exhaustvelocities. Unfortunately this also means that more mass is needed, whichis not preferrable.>This statement bugs me because it is completely contradictory tosomething I worked out a while ago and that you seemed to agree with,which is that higher exhaust velocities are best, and the ideal case isturning all your fuel into zero-mass photons moving at the speed of light.Yes, that was a long time ago, that's why some time ago (2 months) Iasked you if you could find that letter. Since you couldn't find it I starteddigging... deep, finally I found it, but now I lost it.... So dug it up again,Sept 1, 1995 that's the date of your letter. There you say:Choice 3: The fabled Photon Rocket, exhaust velocity r = 1, f = 0.268 Assuming you could make a machine that eats matter and produces astream of pure photons (probably not physically impossible, but still verydifficult) then you get the best fuel efficiency of all, about 3 parts fuelfor each part payload. (note: f = fraction of mass that can be payload) Your definition of fuel-efficiency is completely different from mine(momentum:energy-ratio), your f is the same as the ship:fuel-ratio. >You do want to get all your exhaust products moving backwards, whichimplies confining the fuel reaction and reflecting any forward-movingproducts backwards. You're absolutely going to lose energy to heat if youtry to equalize the exhaust velocity. In any case a gas at a particulartemperature doesn't have a uniform set of particle velocities; it's still astatistical distribution about a mean that is characteristic of the gastemperature.Yes, I'm aware of this heat-loss problem, it was a theoretical model, ofwhich I in the beginning had not thought enough of the practical problems. TimothyP.S. This is something I derived some time ago: Momentum/Energy (p/E) ratio:for photons: 1/cfor mass : 1/c {v/(c-Sqrt[c^2-v^2])} if you know the velocity or 1/c Sqrt[U(U+2 m c^2)] if you know mass and kinetic energy where {v/(c-Sqrt[c^2-v^2])} is always bigger than 1 (for v->c it goes to 1,for v->0 it goes to infinity) So if you have some energy and want to get the most momentum of it youcould best use mass, the efficiency is always better. Note, while the p/E ratio for mass may decrease for increasing v, thisdoes not mean that the total momentum decreases.================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Thu, 06 Jun 1996 23:17:19 +0100To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: Equations page ?>Hey Tim,>I'm working up a new LIT web links page and was going to add yourRelativistic equations page but it doesn't show up. Where did it go? >Kelly>http://www.cpedu.rug.nl/~N0642983/lit/calc.html Its already with steve:http://165.254.130.92/lit/calc.htmTomorrow, I'll put it in another directory, I'll let you know TimDate: Thu, 6 Jun 1996 20:11:25 -0700 (PDT) ================================================================Date: Fri, 7 Jun 96 14:42:42 +0200From: zkulpa@ippt.gov.pl (Zenon Kulpa)To: kgstar@most.fw.hac.comSubject: LIT linksCc: zkulpa@lksu.ippt.gov.pl>From kgstar@most.fw.hac.com Thu Jun 6 17:25:19 1996 >Hey Folks,>I'm working up a new LIT web links page if you have technical or personalpages you'ld like linked to. Send me the HTTP://. See below.._______________________________________________________________ Dr. Zenon Kulpa	|Center of Mechanics,	|	All civilizationsInstitute of Fundamental	|	become either spacefaringTechnological Research	|	or extinct.ul. Swietokrzyska 21	|00-049 Warszawa, POLAND	|	Carl Sagan______________________________|________________________________ tel ++48-22/261281 ext 279	fax ++48-22/269815URL http://www.ippt.gov.pl/~zkulpa_______________________________________________________________Date: Fri, 7 Jun 96 16:37:08 +0200==================================================================From: zkulpa@ippt.gov.pl (Zenon Kulpa)To: kgstar@most.fw.hac.comSubject: Re: LIT links>From kgstar@most.fw.hac.com Fri Jun 7 16:27:10 1996 To:zkulpa@ippt.gov.pl (Zenon Kulpa)That is the proper version of my e-mail... >Cc: zkulpa@lksu.ippt.gov.plThat is the old, improper version (still works, but unofficially - and maycease to work any time).>I still had you at>http://www.ippt.gov.pl/zmit/www/zkulpa/zkulpa.html >I'll change it to>http://www.ippt.gov.pl/~zkulpa>If that is you (or your providers) preference. As above, with e-mail: old version works through alias, but "physically"the new one is currently valid. And it's shorter, too...Regards,-- Zenon================================================================================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Fri, 07 Jun 1996 18:05:12 +0100To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: mind lock http>Ok I have some refs in my web pages such as this: ><p><A HREF="Explorer_Class/">Explorer Class</A> >The idea is you click on the text "Explorer Class". It forwards you to thesubdirectory "Explorer_Class/" and brings up the default document"default.htm". Instead it goves me a directory listing of the contents ofthe subdirectory "Explorer_Class/". Ive tried "default.htm", "default.html"and tried pulling the / out of the Href. NO go. Now I no this worked before,so what did I do wrong?!! Maybe, it is default.html (with L) instead of default.htm (without L) But to solve the problem forever, why don't you just refer to the name ofyour page, instead of relying on a predefined page-name. Yes I know, ifsomeone enters the addres without name, s/he is seeing your dir. In thatcase just add an extra file telling them they entered at the wrong point.================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Fri, 07 Jun 1996 18:05:18 +0100To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: long filenames>DOS! I talk to someone who uses DOS!! I'll be laughed out of cyber space!!<thud>Well, linux isn't my taste, one needs to get all software from the internet,and really I do know only one person who uses it. So then the only alternative is a Mac, well here in the Netherlands almosteveryone has an IBM-based system, so you would be very isolated. I really don't understand why Microsoft never attempted to upgrade the 8-bit DOS to something more mature like 32-bit. >Well I'm bigger on descriptive names than short ones, and normally beingin Macintosh space I forget the constraints of primatives. But I'll try toremember to keep the first chars unique. Won't you get tired from typing the long names? Or do you always usesome kind of scrolling filename list?What is the maximum length of a filename? >>P.S. I know there are programs that rip a whole site (or several levels ofa site), but these aren't the best solution. >rip a whole site?Yeah, you only need to give a starting page, and its starts ripping all linksthat spring from it, and then it starts with the next level: the links fromthe linked pages, and so on. You can give the number of levels you like, anda few other restrictions, so you wouldn't need to be scared that theprogram will rip the whole internet (but if you like, it probably would bepossible).Tim================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Fri, 07 Jun 1996 18:05:15 +0100To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: Equations page ?>>Its already with steve:>Do you mean Dave? Thats Dave's W/S url. Ah yes, sorry, I wrote it in all haste, after I had been looking for an oldletter from Steve for half an hour.>Ok. I guess if its going to be on the site I don't need to link to it anyway. >You equation page is more ledgible now. Stil hard to follow, but easier toread. ;)Once the Java-script is added, it will even be useful for people who don'twant or are not able to work through all the equations. The URL you mentioned last time does no longer exist, the account is gone,but I've a new location for my homepage: (please update your hot-list) http://huizen.dds.nl/~sheliak/welcome.html I already had uploaded my complete calculations-page before you didupload your pages.I had reserved the "lit" directory for my part, but unforunately you used itas a root-directory. (It seems we all share the same space.) So today Imade a new directory for my part and move my stuff there, it is in thedirectory lit/calc, so please do not run over it again. OK, here the full address of my calculations-page: http://165.254.130.92/LIT/calc/calc.html Timothy================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Fri, 7 Jun 1996 11:10:54 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re: mindlock httpCc: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) At 6:05 PM 6/7/96, Timothy van der Linden wrote: >>Ok I have some refs in my web pages such as this: >><p><A HREF="Explorer_Class/">Explorer Class</A> >>The idea is you click on the text "Explorer Class". It forwards you to thesubdirectory "Explorer_Class/" and brings up the default document"default.htm". Instead it goves me a directory listing of the contents ofthe subdirectory "Explorer_Class/". Ive tried "default.htm", "default.html"and tried pulling the / out of the Href. NO go. Now I no this worked before,so what did I do wrong?!! >Maybe, it is default.html (with L) instead of default.htm (without L) >But to solve the problem forever, why don't you just refer to the name ofyour page, instead of relying on a predefined page-name. Yes I know, ifsomeone enters the addres without name, s/he is seeing your dir. In thatcase just add an extra file telling them they entered at the wrong point.I may do that. It just bugs me that it stoped doing something it was doingfine a week ago!Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------================================================================X-Sender: kgstar@pophost.fw.hac.comMime-Version: 1.0Date: Fri, 7 Jun 1996 11:33:11 -0500To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: Re: longfilenamesCc: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) At 6:05 PM 6/7/96, Timothy van der Linden wrote: >>DOS! I talk to someone who uses DOS!! I'll be laughed out of cyber space!!<thud>>Well, linux isn't my taste, one needs to get all software from theinternet, and really I do know only one person who uses it. >So then the only alternative is a Mac, well here in the Netherlands almosteveryone has an IBM-based system, so you would be very isolated. >I really don't understand why Microsoft never attempted to upgrade the8-bit DOS to something more mature like 32-bit. Because they are phasing out DOS. Why upgrade it when you want people todump it?>>Well I'm bigger on descriptive names than short ones, and normally beingin Macintosh space I forget the constraints of primatives. But I'll try toremember to keep the first chars unique. >Won't you get tired from typing the long names? Or do you always usesome kind of scrolling filename list?scrolling lists or icons. You seldom type a file or folders name after youname it.>What is the maximum length of a filename? 31 characters.>>>P.S. I know there are programs that rip a whole site (or several levelsof a site), but these aren't the best solution. >>rip a whole site?>Yeah, you only need to give a starting page, and its starts ripping all linksthat spring from it, and then it starts with the next level: the links fromthe linked pages, and so on. You can give the number of levels you like, anda few other restrictions, so you wouldn't need to be scared that theprogram will rip the whole internet (but if you like, it probably would bepossible).Humm.Kelly---------------------------------------------------------------------- Kelly Starks	Internet: kgstar@most.fw.hac.comSr. Systems EngineerMagnavox Electronic Systems Company(Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------==================================================================Date: Fri, 7 Jun 1996 21:07:15 -0700From: Steve VanDevender <stevev@efn.org> To:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Cc:KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUSubject: Energy:momentum ratioTimothy van der Linden writes:>Your definition of fuel-efficiency is completely different from mine(momentum:energy-ratio), your f is the same as the ship:fuel-ratio. There's one problem here. momentum and energy in conventional units don'thave the same units. Efficiency is by nature dimensionless. mass/mass isdimensionless.>Momentum/Energy (p/E) ratio:>for photons: 1/c>for mass : 1/c {v/(c-Sqrt[c^2-v^2])} if you know the velocity >or 1/c Sqrt[U(U+2 m c^2)] if you know mass and kinetic energy >where {v/(c-Sqrt[c^2-v^2])} is always bigger than 1 (for v->c it goes to 1,for v->0 it goes to infinity) I don't know where you got this "v / (c - sqrt(c^2 - v^2))" bit. How did youderive this? Why do you think it's correct? >So if you have some energy and want to get the most momentum of it youcould best use mass, the efficiency is always better. There's another fundamental flaw in your argument. You can't convertenergy to momentum. Momentum is momentum and energy is energy andone cannot be converted into the other.When you take something that has momentum, and you get energy out of itby slowing it down, you are getting the extra energy it has to compensatefor its momentum (m^2 = E^2 - p^2). If m is constant, then when pincreases E also increases.>Note, while the p/E ratio for mass may decrease for increasing v, thisdoes not mean that the total momentum decreases. In the formalism I'm used to, where v is dimensionless, p/E is exactly thesame as v. If you want to use conventional units then p * c^2 / E is exactlythe same as v.Frankly, I don't know where you're coming from on this and your units don'twork out in any useful way (momentum / energy as a measure ofefficiency has units seconds per meter? What the hell is that?). Efficiencyis dimensionless; it is (energy out) / (energy in).================================================================X-Sender: S9421793@mail.student.utwente.nl Mime-Version: 1.0Date: Sat, 08 Jun 1996 15:20:37 +0100To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,neill@foda.math.usu.edu, 101765.2200@compuserve.com,MLEN3097@Mercury.GC.PeachNet.EDUFrom: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)Subject: Re: USING A ELECTROMAGNETIC BEAM TO PROPELL A STARSHIP >>>Of course if your using my hybrid fuel/sail configuration you want ahavyer sail... ;)>>Laser forth and maser back?>Doesn't really matter. On the way out it depends on what lithium reflectsbest. On the way back probably the same thing, just so you don't need tobuild new beaming facilities.>>>I worry about EM and secoundary reactions in the plasma. We are afterall talking about hellish amounts of power. Even small percentages couldincinerate the engine and ship. (Even a mile of Lithium might not protectthe crew.)>>OK, then we're back on one line...>One line?Eh, I guess my Dutch translation was a bit to litterally: "We're in line with eachother" or "We're in agreement with each other" Timothy