BigMoosie

Maybe if you're crosseyed.

5614: I believe she is using win XP and has told me that two computers in her household fail to run it properly and both are behind the same firewall so think you may have found the possible culprit. I have realised a possible fix however; if instead of using the abstract Image() object I use a real image tag which is hidden then when the onload event triggers I can then slowly display it (incrementing the opacity) over the other image, there should be no problem doing it that way as once it loads for the first time there is no necessity to load it from cache.

She is using IE6. At the moment I have it set up so that the next image is preloaded using new Image(), and as soon as it loads I apply that image in the slideshow, however I think her problem is that after it preloads it is trying to download it for a second time instead of loading it from the cache. I get the exact same problem when I apply this setting in IE: Check for newer versions of stored pages: Every time you visit (temporary internet settings) However changing that on her computer doesn't fix a thing. Is there another setting that causes the same result?

I have built a JavaScript slideshow for a client. It works fine for me but she says there are problems when she views it on her machine. I would be very greatful if you would view the slideshow and tell me whether or not it works correctly. You do not need to be able to understand JavaScript to help me out with this. I just want to know if her problem occurs for alot of people. Please view it here: http://frankipollick.com/slideshow.html and let me know if it works as intended. She claims that the images will be replaced with blackness for unpredictable amounts of time when the slideshow is on. This should never be the case, the images should transition one to the other. Thankyou very much, -Moosie

Sorry for using wrong terminology, and you are correct I do mean the change in the net force. Klaynos, for a more precise question, is the gravity at the poles weaker than the gravity at the equator? Judging from the values you supplied swansont I would assume that it would be. But how about if the moon is directly over-head and if the moon was directly opposite you, what would be the relative differences? And finally how much does the Earth's orbital eccentricity effect it? On a similar note I have heard that there are people who go insane on full moons (lunatics), and that they claim this is because of its gravitational pull, but that doesn't seem to make any sense seeing that the full moon depends on angles, and would be unrelated to its effect on the net force of our gravty. [Tycho?] I couldn't seem to find that image.

I know that the variation in Earth's gravity varies due to the following (and other) reasons: Altitude Variations in the crust density Celestial objects Distance from the equator (centifugal accelleration due to Earth spinning) But which is the most influential and what is their relative magnitudes?

Number of ways to pick k items from n choices. I like it because I solved it myself before I was studying probability, only to learn that Pascal beat me by several hundred years. [math]^nC_k = \frac{n!}{(n-k)!k!}[/math]

Let the porabola be described by: [math]x^2 = 4ay[/math] Where a is the distance from the vertex to the focus. We want a specific region of x-values of this curve lets say from b to -b. That means the equation satisfies (b,a) [math]b^2 = 4a^2[/math] [math]b = 2a[/math] This means the porabola must be 4 times wider than it is tall. This may be enough to solve your problem, but if you need specific dimensions the following is relevant: The length along any curve defined by f(x) is: [math]L = \int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} \cdot dx[/math] So: [math]f(x) = \frac{x^2}{4a}[/math] [math]f'(x) = \frac{x}{2a}[/math] [math]8 = \int_{-b}^b \sqrt{1+\frac{x^2}{4a^2}} \cdot dx[/math] [math]8a = \int_0^{2a} \sqrt{4a^2 + x^2} \cdot dx[/math] let x = 2.a.tan(w) then: dx = 2.a.sec2(w).dw [math]8a = 4a^2 \int_0^\frac{\pi}{4} sec^3(w).dw[/math] [math]a = \frac{2}{\int_0^\frac{\pi}{4} sec^3(w).dw}[/math] I had to use some software here to evaluate this integral, I just couldn't seem to crunch it. [math]a = \frac{4}{log_e(\sqrt{2}+1)+\sqrt{2}} \approx 1.742473598...[/math] So your porabola would fit inside a box measuring 6.9699 x 1.7425 (in feet).

I recently heard of a different kind of implicit differentiation, apparently it works like this: [math]\frac{dy}{dx} = - \frac{dt}{dx} \div \frac{dt}{dy}[/math] Where t is a made-up variable. This makes complete sense to me except for the negative sign. Why is there a negative sign and what is the name of this method?

I have installed a keylogger, but I would use a webcam if I had one, that is a good idea. I fear though that she may not use the computer next time she enters the room because she knows I suspected her last time so a webcam might be the only thing that will work. I will have to see.

Well, there is nothing very important on my computer, I just want to catch her red handed.

Hmmm... I ran a search and found a few files that were opened in the time period. I am pretty certain that it was my sister and not some random program doing an internet update. My main concern is that she had access to my room where I store copious amounts of chocolate and alcohol that I'm sure she must have gotten into. I think I will install a more advanced logger, know of any?

I have read that using Newtonian physics, the gravitation interaction between two bodies traces the path of a conic, if one is orbiting the other it is an ellipse, if it is a fly by it forms part of a hyperbola, somewhere in-between is a porabola. It seems that a simple condition for two bodies to be orbiting is that one body needs to make a 180 degree turn, if that is so then it could only possibly be an ellipse and therefore be in orbit. Taking relativity into account, is it possible for two bodies to come near each other and one make a 180 degree turn but still end up being ejected from the field?

I don't quite understand your graph but perhaps it answers this question. It seems to me that a ray of light would only be able to orbit the black hole at an exact radius, any less and it would hit, any greater and it would stray (presumably this radius = the event horizon?). So it would be an extremely thin shell of light that orbits the black hole, if the black hole were to change mass in the slightest then all of the trapped light would end up leaving this radius and thus there would never be a great deal of it at any given time. Or perhaps there is a range of radii and this is all wrong.

Hi, I came home and opened winamp today to find a file in the playlist that I have not opened recently, I checked the winamp.ini file and discoverd that it was last modified at 8:08pm, a time that I was out of the house. I am interested in finding out what was accessed on my computer, there are no strange cookies or temporary internet data but what else can I check and how? Regards, Moosie

My mum today brought home a dozen free range eggs and strangely every single one contained two yolks inside it. I found this to be quite odd, would they most likely all have come from the same hen? Is it drugs?

Thanks Betty and Swansont!

I love this little tune but do not know where it is from, do you know anything about it? Download MP3 (200kB) This is only a 20 second extract. Cheers!

The problem is resolved but I am still curious about something, is the absolute sign that you suggested related to the definition of logarithms in the complex plane?: [math]log_e(z) = log_e(|z|) + arg(z)i[/math] If so, why is it that the improper integration you mentioned does not work out like so? Does it make sense to have complex area? [math]\int_{-1}^{2} \frac{1}{x} dx = [log_e x]_{-1}^{2} = log_e(2) - log_e(-1) = log_e(2) - (log_e(1) + \pi i) = log_e(2) - \pi i[/math]

Thankyou very much woelen, that answers my question perfectly.

I am working on some problems on particle dynamics and I keep making errors originating from the following: [math]\int \frac{1}{x} = log_e x[/math] or [math]\int \frac{1}{x} = \int \frac{-1}{-x} = log_e -x[/math] Why is it that there are two solutions and how can I identify which is the correct one to follow?

Hi, I have recently purchased a PDA which does not support WMV video formats. I was wondering what suitable software would be able to convert these to either MPG or AVI files while keeping the file sizes similar. I obviously do not need something overblown like Adobe Premier, but something with a few customise options such as quality and resolution etc would be good. Does anybody know of any such free program? Thankyou very much! -Moosie

I have the open office installer on my computer, I just didn't realise that it was capable of doing that. Thankyou everybody. whap2005, thanks, I'll see how OO works out and if I need more I might catch you up on that.

Does anybody know any programs that are able to create PDF files? I cannot seem to create them with Acrobat. I can find all sorts of programs with a Google search but I would like to know what is recommended. Preferrably a free program. Thanks.

Would it be possible to solve the volume from a CT scan?