Reply to thread

Message

<blockquote data-quote="freyar" data-source="post: 7494587" data-attributes="member: 40227">Just a reminder, we are looking for solutions to the equation 1=cos(x)+0.0602 sin(x) between x=0 and 90 degrees.&nbsp; I suspect there is some problem with your excel, and I'll see if I can attach a plot.&nbsp; But you are not limited to excel.&nbsp; Try <a href="https://www.wolframalpha.com/" target="_blank">Wolfram alpha</a> (free!).&nbsp; If you type cos(x)+0.0602 sin(x)=1 in the box, it will give you a plot showing both solutions in the 0 to 90 degree range as red dots.&nbsp; It will also give you a formula for all solutions below that.&nbsp; If you want numerical values, you can click the &quot;approximate solution&quot; button above the formula to get the answer in radians, which convert to degrees by multiplying with 180/pi.&nbsp; If you want to have a zoom in of the plot, you can instead type &quot;plot cos(x)+0.0602 sin(x) from 0 to 10 degrees&quot; (or whatever range you want) in the text box.&nbsp; But it should be clear from the first plot that the only solutions between 0 and 90 degrees are x=0 and the one at about 0.12 radians.Here are my plots. The blue curve is our trig function, and the purplish line is 1.&nbsp; The horizontal axis is the angle x in degrees.&nbsp; You can see there is a solution at x=0 and 6.89 degrees.&nbsp; Incidentally, I'm using Mathematica for plotting today, since I like its plotting features better.[ATTACH]101470[/ATTACH]This comes back to my original formula.&nbsp; We agree that the angle of deflection is tan(a)=dp cos(x)/( p-dp sin(x))= (dp/p) cos(x)/(1-(dp/p)sin(x)).&nbsp; We were talking about an example with dp/p = 1/2.&nbsp; I claim that the maximum deflection --- where tan(a) is biggest --- occurs for sin(x)=1/2, which is x=30 degrees.&nbsp; I also claim that maximum deflection is given by tan(a)=sqrt(3)/3.&nbsp; Here is my plot.&nbsp; The blue curve is tan(a) as a function of x (horizontal axis in degrees), the purplish line is sqrt(3)/3=0.57735..., and the red vertical line is at 30 degrees.&nbsp; You can make your own judgment about whether my formula is correct.&nbsp; You can also do a version of this plot on Wolfram alpha for yourself, but you'll probably have to use radians for x.[ATTACH]101472[/ATTACH]Editing because it posted when it was supposed to preview: Anyway, we can also figure out the minimum value of dp/p needed to deflect the asteroid by required angle given by tan(m).&nbsp; We just use the optimal angle given by sin(x)=dp/p for a given dp.&nbsp; Then the deflection angle (the best one for that dp) is tan(a) = (dp/p) sqrt(1-(dp/p)^2)/(1-(dp/p)^2) = (dp/p)/sqrt(1-(dp/p)^2).&nbsp; Then you have to find when this tan(a)&gt;=tan(m).&nbsp; I believe that equality occurs at dp/p=sin(m), so that should be the minimal value of dp/p that will deflect the asteroid the required amount.Once again, I've been assuming dp/p&lt;1 because otherwise Pierce could just say, &quot;this is simple,&quot; and just be a dumb brute and blast the asteroid back from whence it came without having to stress or think about it.</blockquote>

[QUOTE="freyar, post: 7494587, member: 40227"] Just a reminder, we are looking for solutions to the equation 1=cos(x)+0.0602 sin(x) between x=0 and 90 degrees. I suspect there is some problem with your excel, and I'll see if I can attach a plot. But you are not limited to excel. Try [url=https://www.wolframalpha.com/]Wolfram alpha[/url] (free!). If you type cos(x)+0.0602 sin(x)=1 in the box, it will give you a plot showing both solutions in the 0 to 90 degree range as red dots. It will also give you a formula for all solutions below that. If you want numerical values, you can click the "approximate solution" button above the formula to get the answer in radians, which convert to degrees by multiplying with 180/pi. If you want to have a zoom in of the plot, you can instead type "plot cos(x)+0.0602 sin(x) from 0 to 10 degrees" (or whatever range you want) in the text box. But it should be clear from the first plot that the only solutions between 0 and 90 degrees are x=0 and the one at about 0.12 radians. Here are my plots. The blue curve is our trig function, and the purplish line is 1. The horizontal axis is the angle x in degrees. You can see there is a solution at x=0 and 6.89 degrees. Incidentally, I'm using Mathematica for plotting today, since I like its plotting features better. [ATTACH=CONFIG]101470._xfImport[/ATTACH] This comes back to my original formula. We agree that the angle of deflection is tan(a)=dp cos(x)/( p-dp sin(x))= (dp/p) cos(x)/(1-(dp/p)sin(x)). We were talking about an example with dp/p = 1/2. I claim that the maximum deflection --- where tan(a) is biggest --- occurs for sin(x)=1/2, which is x=30 degrees. I also claim that maximum deflection is given by tan(a)=sqrt(3)/3. Here is my plot. The blue curve is tan(a) as a function of x (horizontal axis in degrees), the purplish line is sqrt(3)/3=0.57735..., and the red vertical line is at 30 degrees. You can make your own judgment about whether my formula is correct. You can also do a version of this plot on Wolfram alpha for yourself, but you'll probably have to use radians for x. [ATTACH=CONFIG]101472._xfImport[/ATTACH] Editing because it posted when it was supposed to preview: Anyway, we can also figure out the minimum value of dp/p needed to deflect the asteroid by required angle given by tan(m). We just use the optimal angle given by sin(x)=dp/p for a given dp. Then the deflection angle (the best one for that dp) is tan(a) = (dp/p) sqrt(1-(dp/p)^2)/(1-(dp/p)^2) = (dp/p)/sqrt(1-(dp/p)^2). Then you have to find when this tan(a)>=tan(m). I believe that equality occurs at dp/p=sin(m), so that should be the minimal value of dp/p that will deflect the asteroid the required amount. Once again, I've been assuming dp/p<1 because otherwise Pierce could just say, "this is simple," and just be a dumb brute and blast the asteroid back from whence it came without having to stress or think about it. [/QUOTE]

Verification