Thanks, Charles :D

Physics and a few jumps

What fun can we have with Physics?

Submitted By Mars Explorer, Quantum, Community

This Document originally posted in the "Mars Explorer" Group
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

d=distance

t=time

Vox=the original x velocity

a=Acceleration, or gravity

y=up and down, or vertical

x=left and right, or horizontal

V=velocity

^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 :)

» Reply to CommentRe: Physics and a few jumps

Thanks, Charles :D

» Reply to CommentRe: Physics and a few jumps

I was also thinking of using this to help out racing techniques, such as in GTR's Desert. You know that first hill that you go down? Well, if you knew how much you had to slow down, and you wanted to go as fast as possible (especially in a hovercraft race), one thing you could do is use triangles.

What you would do to measure these things such as hill falls would be to have the quarry go to the top of the hill, or the part where you start to go off, then go to the part where you want to land, look at the DST, band because it works in 3 dimensions, it will give you the Distance of the straight line to the buggy, not just the x direction distance, I think. Now, that distance would be the hypotenuse. to find the height, you would find the differnce of altitude between you and the quarry. You would do this just by looking at your altitude before you go down the hill, and then compare it to the altitude at the bottom of the hill. Now, using pythagorean theorum, a^2=b^2=c^2, you can find the base, or the distance of the fall in the x direction. Lets say the hypotenuse of the jump was 100 meters, and the height was 60. 3600+x=10000. The square root of 6400 is 80! That would mean that the base of the jump was 80 meters.

Now, you can find how fast you want to go maybe by using the time to fall equation, which is t=square root of 2(Y)/a, with a being in the square root. All of this gets into details, but maybe when I have the results if I do this, I will tell you how I did these things for all of these jumps, falls, and launches :)

Edit: Should I delete this and add it on to the original thread or just keep it here as is?

» Reply to CommentRe: Physics and a few jumps

Oh my god...there is ony a few words to say this..but...Wtf is this?

» Reply to CommentRe: Physics and a few jumps

This is taking advantage of the physics of the game to increase how much we enjoy it by awareness :)

It's also math and trigonometry...

» Reply to CommentRe: Physics and a few jumps

Math is awesome.

Physics is awesome.

You're awesome for making this post.

» Reply to CommentRe: Physics and a few jumps

Watch out, when Yoshi gets here, he will love this post like no one can even comprehend :P

» Reply to CommentRe: Physics and a few jumps

I am the almighty CharlesHadouken.

I'mma read this again and again and again.... *infinity*

» Reply to CommentRe: Physics and a few jumps

Lol............................

» Reply to CommentRe: Physics and a few jumps

Well, in the custom server settings list, gravity does show up as 9.81, therefor, mars gravity is very similar to real life.

» Reply to CommentRe: Physics and a few jumps

Yes, I just thought it was a bit cool.

Lol, however, if we were only to have natural gravity instead of gravity that we can change, there would be very little at all, because an island does not have that much mass relative to say a planet.

» Reply to CommentRe: Physics and a few jumps

Yeah, but 9.81 what? Green fart lengths?

With this he proved that DST is in meters.

» Reply to CommentRe: Physics and a few jumps

(duplicate)

» Reply to CommentRe: Physics and a few jumps

I read this post again to understand it to only conclude that i have better chances riding my unicorn across a rainbow.

» Reply to CommentRe: Physics and a few jumps

We could measure the sky king and compare it to a star destroyer >:3

» Reply to CommentRe: Physics and a few jumps

Guys, update for Bagel's Skate Park Loop-de-loop experiment!

» Reply to CommentRe: Physics and a few jumps

I'll try it at 46 mph! :)

» Reply to CommentRe: Physics and a few jumps

One thing, though, this speed is only for the TOP of the Loop, where the most G forces are needed, or where the required centreptial force is 1G relative to the Gravity on the server. There are other required speeds for other parts of the loop.

» Reply to CommentRe: Physics and a few jumps

Guys, I am trying to attempt something, but I need people's help to do it. I am trying to measure exactly how long GTR's Desert track is. This would be fun to know for the racing fans out there. With the knowledge of the length of the track, you could figure out your average speeds along with the lap times which you already have access to. And, for those wanting a much longer race, you could actually say that this is the GTR's Desert 100 or 500, instead of just naming it that for fun.

What I already did, just for a rough estimate, was I tried to drive around GTR's Desert Track at a steady 60 mph. Now, doing that is very difficult, and there were some changes in speed, especially on the bumpy parts and elevation changes. Now, at 60 mph, you are going a mile a minute. At that speed, it took me 11 minutes and 32 seconds to go around, so maybe a close rough estimate would be that the GTR's Desert Track is a whopping 11-12 miles long!

What I already started doing was having someone be quarry, and I measure certain lengths, and add them up. I already got done maybe almost a mile of the track measuring the accurate way, so it will take a long time, maybe even an hour or 2 cumulatively. I hope I can get this done, and maybe evn for Dvo's Oval track on Dvo's Stunt Parkway, because knowledge of these things gives us so much more to do with our marsxplr time.

» Reply to CommentRe: Physics and a few jumps

12 miles! Wow!

I should go mesure Pancakes' race...

I should go mesure Pancakes' race...

» Reply to CommentRe: Physics and a few jumps

Brilliant! You have my congratulations :)

» Reply to CommentRe: Physics and a few jumps

Thanks! This is just what I do for enjoyment, so it wasn't really work to do it :D Plus, it's nice to have people's support to further that motivation.

» Reply to CommentRe: Physics and a few jumps

Oh, and your mathematical merit has earned you a slot in the Directory of Mars!

congratulations,

ffroggy

» Reply to CommentRe: Physics and a few jumps

Lol, directory of Mars? forgive me if I don't know exactly what that is, and if it pertains to marsxplr or the actual planet of Mars :D

EDIT: Oh! Now I see it in the forum :) Thanks!

» Reply to CommentRe: Physics and a few jumps

All that math made head explode into candy. My siblings are picking it up off the floor. HOW DO YOU DO THIS CRAZY MATH?!?! IT'S AMAZING!!!

» Reply to CommentRe: Physics and a few jumps

The gravity graphics are amazing because they are exact. 9.81 m/s^2 was entered into the code for mars.

Now what you forgot to do was run trials. Obviously, you had error because you assumed. So you need to run trails, about 3.

Good luck! :)

(CharlesHadouken)