I was playing Marsxplr today and told the Quarry, the almighty CharlesHadouken to stay still in the spawn area of Freestyle. I thought that 1 DST was 1 meter, so I wanted to do a test. I drove up to that bump you see in front on the Spawn area, and hit it at 89 mph. It was at 150 DST from the Quarry, and I landed at 215 DST, about.
I thought this was pretty cool to test out and see how accurate the default gravity was to Earth's gravity, and it is remarkable :)
To prove this, I popped up a google images photo of a protractor and estimated that my jump was at 12 degrees.
I went on to wolfram alpha (Time to do trigonometry and conversions!), and I put in the numbers. I used a projectile equation, 2(Voy/a)(Vox), with Voy being the original Y velocity, and Vox being the original X velocity.
Ok, 89 mph is 39.79 m/s, so that is the Vo total. 39.79sin12 is the Voy, which was 8.27281,
39.79cos12 is the Vox, which was 38.9205. All good.
2(8.18548/9.807)(38.5097)=65.6366 meters :) Very close! As the original estimate was 65 meters travelled in the X direction. Just to throw it out there for guys who don't know it, the Gravity constant of earth is 9.807 m/s^2
Now, my point here wasn't just to prove how amazing the gravity physics of this game are, we already know that.
What I did want to do with this was have a little fun.
Maybe, we could figure out all the required speeds for certain jumps and loop-de-loops were for all of the skate parks. I mean of course, not all of us like to think of the science behind it, that would be a bother if we did it all the time. What is fun though, is awareness of the game, that is how you get the full experience, thinking about how realistic the experience is, and maybe even using is to better your skills or something similar to that :)
We could also test out the other gravity settings too, like what m/s^2 is full.
Sorry if someone has made up this idea before, I just thought that this would be fun to write :)
Thanks, everyone, for absorbing this useful and interesting information :D
EDIT: Now, this is the section where I show all of my results, though they are not all coming at once, I still need to do a lot of work.
Just to give you an overview of how I work these out, I use any of these equations to work out a problem. But sometimes, I may think of other, new ones for different situations, so I will add them here if I do.
Voy=Original y velocity, usually used during projectile equations when there is an angle of elevation of the original jump
Vox=the original x velocity
a=Acceleration, or gravity
y=up and down, or vertical
x=left and right, or horizontal
^2=Squared, so ^ and a number is to the power of a certain number
r=radius, or half of the diameter, or 6.28... times less than the circumfrence of a circle
2(Voy/a)(Vox)=Total distance travelled for a projectile
(Voy/a)(Vox)=X distance at Y max during the projectile motion
(Voy/a)(Vy average)=Maximum height reached during projectile motion
t=Root 2(y)/a with everything in the square root=Time for something to fall from a certain height stating with 0 velocity
t=root 2(y)/AsinTheta=Gravitational acceleration down a slope, with theta being the angle of elevation of the slope or ramp, and again, everything is in the square root for that equation, too
V^2/r=Centripetal force of turning in a circle or curve at a speed
Bagel's Skate Park Loop-de-loop
I'm sure all of you know about Bagel's Skate Park, and the loop-de-loop that is in it. Now, of course, you have figured out that you cant go around the loop-de-loop at 10 mph without falling off of it. There is a way to find the minimum required speed. Centripetal forces are measured with the equation V^2/r... This equation means velocity Squared divided by the radius of the turn is the m/s^2, or the g forces. 9.807 m/s^2 is 1 G. Now, intuition would tell us that we would need 1 G of acceleration at the top of the loop-de-loop in order for us to keep on driving on it without falling down, or at least without our wheels letting go of it (You probably could drive at lower speeds and cut across the top of it, but if you drive it too slowly, you may tumble over from the impact).
To figure out the minimum required speed, all it took was some easy math. I flew to the top of the loop-de-loop and drove around a bit to find it's highest point. That was at 98 altitude, or 98 meters high. When I drove to the bottom, I drove around a bit and found out that it's lowest point is at 13 meters altitude. That means that the diameter of the loop-de-loop is about 85 meters. Dividing it by 2 would give us the radius of 42.5 meters.
To find out what speed met the 9.807 m/s^2 criteria, I just multiplied 42.5 by 9.807 and got 416.7975. Now, that 416.7975 m/s was the velocity, but squared, so, taking the square root of that, you would get a velocity of 20.41562 m/s. Now, if you converted that to mph, you would get a minimum required speed of around 45.6683 mph...
Just a quick fact about full gravity loop driving, guys, at the full gravity constant of 18.81 m/s^2, it would take a minimum of 63.247 mph. However, that is very difficult to reach with the full gravity in the way, so I suggest starting out with a default gravity, and then turning it to full once you have picked up enough speed to do a full loop.
I hope that was interesting for you guys :)