c3k

Slaphappy and Splinty, You are both, of course, correct. The nvlddmkm issue is a video driver issue that both nvidia and ati (ati's has a different prefix, atiddmkm?) problem that they, the video manufacturers and microsoft, refuse to acknowledge. My criticism of BF.C in this regard is unfounded. My apologies. Regards, Ken

handihoc, What do you mean about all those new scenarios and campaigns? I have downloaded v1.07 and only have the two campaigns (Training and Thunder) and the same 19 scenarios. Did you get more with the patch? If not, where did you find them? Thanks, Ken

Melnibone, Thanks, but, oddly, that file does not exist. Within the "Game Files" folder all I have are these three: Campaigns, Quick Battle Maps, and Scenarios. This is despite having played a v1.07 game from scratch and saving it. The saved game DOES show in cmsf menu. This is odd. Any other ideas? Thanks, Ken

Gents, I have a surplus of saved game files showing in CMSF's menus. I want to delete a bunch of them. I am unable to do so directly through the menu. I am unable to do so using the scenario editor. I assume I will be able to delete them using windows and browsing to the directory. Of course, that also assumes that I could FIND the saved game files. To what directory are saved games stored? I have left all CMSF install options to default settings. Thanks, Ken

Gents, v1.07, BFC version: 18 seconds into my first WEGO turn the same crash has occurred. 'display driver stopped responding: nvlddmkm" No other games have issues on my machine. Vista Ultimate, x64 q6600 8800gtx 8Gb G Skill Let me know when the game works. Regards, Ken

Isn't this situation similar to the attempted secession of 1861 in the U.S.? Food for thought. Regards, Ken

Hey! I played a scenario called "Cowboys and Indians" in CMBO but it was TOTALLY unbalanced. Suddenly some some guy used an uber-panther against my "HQ". See, I'd met this chick in a bar and we went back to her place. Next thing I knew, I had a new scenario on my HARD-DRIVE. Then, that damn panther showed up. Now it all makes sense.... Cheers, Ken

Althought it may seem that Charles should get lauded for finding, and apparently fixing, this CTD bug, that would be wrong. As thread-owner, all laurels belong piled at my feet. Commence laurel throwing. Thank you. Regards, Ken

Where does one find the error page Other Means has posted, above? I should have lots of those files by now. Regards, Ken

Anyone (BF.C) want the file?

No, it's all BF.C version (game, patch). Thanks, Ken

Gents, I have a savegame file with which I can crash my computer every single time. I patched to v1.06 and started a new game. The savegame is 2 minutes into it. WEGO style. The freeze followed by crash occurs less than a second into the execute. Is there anyone interested in getting a copy of this savegame and seeing if it crashes on their machine? If it doesn't, obviously my machine is the issue. If it does crash on another machine, it would help to narrow down a software cause. My specs: q6600 processor, 8800GTX graphics, vista ultimate 64. Thanks, Ken

Steve, The spacebar in v1.06 is great. Thanks for including that in the latest update. Regards, Ken

Same problems here. Hang ups, redraws, stuttering in game, followed by some sort of shutdown message about a C++ runtime error. Then I'll get a "nvlddmkm" error message. Vista x64, evga 8800gtx. Using 169.28. The black screen of death is a sure thing with CMSF. This makes it unplayable. Any thoughts? Thanks, Ken

Great patch! 1.07, Ken

Quad core here. Vista 64-bit allows me to use all 8 gigs of ram. The only issue left is I'm not able to play CMx1 on this machine. Otherwise, it's fine. Ken

BAH! A game company with a set of balls would release a patch, as anticipated as this one is, at precisely the moment of the Superbowl kickoff. That would induce a bandwidth surge which would knock the game off-air, causing wailing, gnashing of teeth, riots, downward spiraling GDP's, recession, and a world-depression. C'mon BF.C, grow a pair: release it on Sunday... Ken

Hmmm, a RED herring? You mean there NEVER was a v1.06? 1.05 is final? BF.C has been using the code name "v1.06" to refer to their dinosaur battle game, soon to be released? This revelation has TOTALLY destroyed my faith in BF.C! aaaaaaaaaaaaaaaarrrrrrrrrrrggggggggggghhhhhhhhhhh!!!!! This is horrible... Later, Ken

I did it first! Look, we all know that SOMEONE is gonna complain about v1.06. This gives them the place to do so. Regards, Ken

I'm just glad that BF.C is working on improving the UI by adding functionality for those who aren't comfortable with, or used to, hotkeys. Thanks, Ken

YankeeDog, Thanks! You are correct. Silly me; I don't even have the excuse that I was drinking at the time. I shall rectify both the decimal placement AND the lack of drinking... Carry on! Regards, Ken

Well, in the maths I did, above, I assumed the shot was fired from the equator directly to north. Another assumption was that there was no effect from any type of crosswind, any deceleration due to drag, bullet rotation, tumbling, etc. Assume the earth is a perfect sphere. The rotational period at the equator is precisely 24 hours and the velocity is precisely 1,000 mph. After the bullet leaves the barrel it maintains the rotational velocity imparted upon it, the gun, the firer, and the ground beneath the firer due to the equatorial rotation rate of 1,000mph. That velocity has a direction pointing directly to the east. Assume the target was directly in sight, and the gun was aimed directly towards the target. The target, being north of the equator, will rotate a lesser distance in the 24 hours per rotation. The rotational velocities I found were: Equator = 1,466.6666666666666667 ft/sec 4 miles north of Equator = 1,466.66586 ft/sec (rotational period at equator is 24 hours; velocity of 1,000 mph, therefore equatorial circumference is 24,000 miles. C=2pi*r. Pi=22/7, solve for r. r(equator)= 3,818.1818181818 miles. Next, the unknown r(4 miles north of equator) must be solved. Given r(equator), and a known distance 4 miles to the north. Approximate the angle between these two lines: from earth center to target point 4 miles north of equator; from target point 4 miles north of equator to firing point on equator: assume this is a 90 degree angle. I told you this is rough! That gives, using Pythagoras's theorum, r(target)= 3.818.1797 miles. The angle at earth center between firer point and target point is solved to be .028648 degrees. With the given radius(target), the circumference is given by C=2pi*r. C=23,999.9867 miles. That, divided by a period of 24 hours per revolution gives the target point velocity.) Therefore, if the bullet maintains its rotational velocity, it will have a greater lateral velocity than the target (given equal gravitational forces and no drag, etc...). That extra lateral velocity is approximately .0008067 ft/sec directly East. Now, compute the flight time of the bullet. I used a constant 1,000 ft/sec. I know; that is a horrid approximation for a 4 mile shot. That 4 mile shot (5,280 ft/mile * 4 miles = 21,120 ft) would take 21.12 seconds. Now it's simply 21.12 seconds * .0008067 ft/sec = .0017037 ft. Or, .20 inches. That's for 4 miles. Again, taking that approximation and multiplying by 1/16 (1/4th distance = 1/4 flight time; 1/4th distance = 1/4 lateral velocity difference; distance = velocity * times; 1 mile distance = 1/4 * 1/4 4 mile distance)gives the approximation for 1 mile; .0125". Obviously, this is the lateral translation to the East. Again, let us know what results are given by experimentation. Thanks, Ken edited to fix my decimal placement errors! [ January 29, 2008, 09:51 AM: Message edited by: c3k ]

Hmmm, I did some very rough approximations. Over a hypothetical 4 mile shot, using an average velocity of 1,000 ft/sec (so about 21.12 seconds of bullet flight time), the coriolis effect would induce the bullet to hit .01703 ft (.20 inches) to the spinward (East) of a target directly North/South of a firer on the equator. That was more than I expected, even given exagerated distances. Using those approximations, and reducing the flight time/distance to 1/4 (1 mile), the velocity difference at latitude would also be very roughly 1/4 the original, so would give 1/16 the original difference. 20" x 1/16 = .0125". All very rough with a lot of initial conditions and only first order approximations, etc., etc. I would recommend a series of actual shots. Let us know what you find out. Regards, Ken edited to fix my decimal error [ January 29, 2008, 09:52 AM: Message edited by: c3k ]

I agree with the breaking the illusion sentiment. If the fight seems fairly conducted, I accept the outcome. If something breaks the illusion, I'll repeat it and counter whatever broke it. But it'd have to be somewhat egregious. Reinforcements attacking me from a secured sector, AI bug resulting in losses - NOT the AI doing something I didn't like. Regards, Ken

zero g, full-displacement drive, null cannon, nano-armored space-marines against angry space-lobsters. Ken