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 [quote="Tom"]I'll have a go at this. Remember force = mass*acceleration.(F=ma). a=dv/dt (deriv of velocity wrt time) v=dx/dt (deriv of position) ax, vx, x is left to right. ay, vy, y up to down Imagine your cannonball midflight. Left-right, no forces at all. So m*ax=0. So ax=0. So vx=cx, a constant. So x = cx*t + dx (another constant). Up/down. m*ay = -g (gravity) So vy = -(g/m)*t + cy. So y = -(g/m)*t^2 / 2 + cy*t + dy. Now, at t=0, y=0, x=0. So dy=0, dx=0. g=10, say, m=100, so our equations are - x=cx*t y=-t^2 / 20 + cy*t Lets lose time. t=x/cx, so subst in, y = -x^2/(20*cx^2) + x*cy/cx. When x=100, we want y=0, so 5 = cy*cx But cy and cx are the original x and y components of velocity, so the orginal velocity is sqrt(cy^2 + cx^2). Now; this is all we know, as you have not said what the angle of the cannon is, and this would vary the velocities. Say a 45 degree angle, then cy=cx, so we get (from 5=cy*cx) cx=sqrt(5), then the velocity is sqrt(10) (from sqrt(cy^2 + cx^2)). Does that make any sense at all?[/quote]
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Ghost Post
Posted: Tue Jul 03, 2001 10:07 pm    Post subject: 1

BTW firemeboy I think your games for Physic demonstration is a great idea. I would love to have something like that in my freshman Physic class (last semester) instead of the professor showing us chart of blocks on a table or balls hanging from ropes.

I have to give him some credit though, he brought it a miniture cannon and shot it at a monkey hanging from an magnet. He had a switch that shot the cannon ball and released the monkey at the same time to prove to us that regardless of the horizontal direction, both monkey and ball will fall at the same speed. The class cheered when that ball smak the monkey in the forehead.

Ghost Post
Posted: Tue Jul 03, 2001 9:56 pm    Post subject: 0

I think the Mafia games is great for kids in Elementary school. Still uncertain on the rules though, great deduction skill seems to be required. Maybe the violence maybe a bit unappropriate ... soften it with alternative wording besides "lynching" and "mafia scums"?
Ghost Post
Posted: Thu Jun 28, 2001 2:59 pm    Post subject: -1

What about the density of the cannon ball or have I lost my mind? Once you fire it from the cannon, at whatever angle you deem most efficient, a larger density (and thus a smaller volume) will travel farther.
Or... are all cannon balls the same size? It's been a while since I've launched anything.
CrystyB
Posted: Fri Jun 01, 2001 11:38 am    Post subject: -2

To me it sounds like Scorch(ed Earth). But i can't imagine how can you teach physics. You could have an information panel with horizontal distance, vertical distance, heights of obstacles between cannons, wind speed, wind direction, and so on? And perhaps provide some blackboard that can be used to calculate stuff and take decision accordingly?

Yours,
Cristian.
mole
Posted: Fri May 25, 2001 10:02 pm    Post subject: -3

I'll play!

or I could make one...
CzarJ
Posted: Fri May 25, 2001 4:18 pm    Post subject: -4

Here's an idea. How bout you make a game where this little kid with a water gun and a finite stash of water balloons takes on the evil fire wizard. You could call it "Timmy in Evil Fire Land."

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Agamemnon
Posted: Fri May 25, 2001 9:57 am    Post subject: -5

firemeboy,
if you need something to fire out of the cannon just let me know, I'll have a few things shipped out to you.
Ghost Post
Posted: Fri May 25, 2001 5:51 am    Post subject: -6

Ok, here's my best explaination of the formulas I used for projectile motion.

a) x = vx*t, where x is the horizontal displacement (distance), vx (the x should be subscript) is initial horizontal velocity, and t is time.

b) vx = v*cosŲ, where v is the initial velocity (on an angle) and Ų is the angle.

c)y = vy*t + 1/2gt^2, where y is the vertical displacement, vy is the initial vertical velocity, and g is the acceleration due to gravity on the earth's surface, approximately 9.8 m/s^2.

d) vy = v*cosŲ

If you understand the concept of two-dimensional vectors, this should not be a problem. If not, I suggest you take a physics course If you want, I can make an example with a diagram, cause those can help a lot.
Mercuria
Posted: Thu Sep 07, 2000 11:47 pm    Post subject: -7

tom: it's not so much that my way is simpler than yours. i just use a simpler problem. with basic physics (and most educational simulations i have seen), you skip figuring out the velocity (impulse and all that... you just leave it out and pretend that you can control the velocity with your magic turny-dial-thing ;-> ), and go straight for the velocity components.
Tom
Posted: Thu Sep 07, 2000 3:20 pm    Post subject: -8

I'm sure my working could be made more elegant, but I don't see how you could do it much more simply. Show us then, Mercuria. (I must admit it's been a good 6 years since I did any mechanics).
Quailman
Posted: Wed Sep 06, 2000 9:34 pm    Post subject: -9

FMB: The trajectory (path) of the projectile (arrow) consists of two components - Horizontal and Vertical.

Horizontal velocity is constant (because we are ignoring wind resistance, drag, etc) and is at most the absolute velocity of the arrow (if you aimed it horizontally). It can be as low as zero if you aim it straight up, but if you do this you'd best move before it comes down.

Vertical velocity depends on two things - the angle at which you aim the arrow and gravity. Because gravity exerts a constant force on the arrow, its vertical velocity is constantly changing.

Before gravity kicks in, the H and V components are figured in relation to the angle you aim. If you shoot at a 45degree angle, they would be equal. The formulas above with Sin and Cos are used to compute this.

Gravity applies to the V component at a constant rate, causing upward velocity to decrease to zero and then become negative as the arrow begins to fall. At a certain point the height of the arrow will equal the height from which you launched it.

You need to do the math to figure out an angle that will impart a certain Horizontal velocity to the arrow such that it covers the Horizontal distance to the target in a given time (T), such that gravity will cause the arrow to return to its original Vertical height in that same time. That, or take a big quiver with you and use trial and error.

Good luck!
daniel801
Posted: Wed Sep 06, 2000 9:33 pm    Post subject: -10

sounds like gorillas.bas from qbasic http://qbplace.hypermart.net/cgi-bin/load.pl?../games/gorillas.zip

great idea to soup it up

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firemeboy
Posted: Wed Sep 06, 2000 8:59 pm    Post subject: -11

Ok, you are probably feeling like you are talking to a wall. Let us assume that I am an archer. I have a bow and arrows so the mass of the arrow is a constant, and the velocity (am I using the correct word here?) is also set. A 90 pound bow will shoot at the same strength every time (I know, not in real life but this is happy-happy land). So the only way the archer can control the distance is by the angle with which he/she shoots the arrow. Is this correct? Where do I go from here.

We can pretend the arrow is 5 grams... How do we figure the constant of the bow, and then the distance and angle.

I feel like I am in over my head here.
Mercuria
Posted: Tue Sep 05, 2000 10:04 pm    Post subject: -12

lol... when'd we bring force into this matter? (yeah, yeah, bad pun, i know ;-> ) you're making it so much more complicated than it has to be...

fmb: you didn't give enough info... there are two variables left... angle and velocity. mass doesn't matter (unless you, as tom suggests, use force instead of velocity... but then a lot more variables and a lot more equations are involved...). you have to specify the angle if you want the velocity.

for horizontal velocity component, let the angle from horizontal be uh... does this (Ų) show up as theta?

anyway, whatever it looks like to you...

horizontal component = v cosŲ
vertical component = v sinŲ

this is all, of course, in the "perfect physics world" with no wind resistance and stuff like that ;->
Tom
Posted: Tue Sep 05, 2000 3:17 pm    Post subject: -13

I'll have a go at this. Remember
force = mass*acceleration.(F=ma).
a=dv/dt (deriv of velocity wrt time)
v=dx/dt (deriv of position)

ax, vx, x is left to right. ay, vy, y up to down

Imagine your cannonball midflight. Left-right, no forces at all. So m*ax=0. So ax=0. So vx=cx, a constant. So x = cx*t + dx (another constant).

Up/down. m*ay = -g (gravity) So vy = -(g/m)*t + cy. So y = -(g/m)*t^2 / 2 + cy*t + dy.

Now, at t=0, y=0, x=0. So dy=0, dx=0. g=10, say, m=100, so our equations are -

x=cx*t
y=-t^2 / 20 + cy*t

Lets lose time. t=x/cx, so subst in,

y = -x^2/(20*cx^2) + x*cy/cx.

When x=100, we want y=0, so

5 = cy*cx

But cy and cx are the original x and y components of velocity, so the orginal velocity is sqrt(cy^2 + cx^2).

Now; this is all we know, as you have not said what the angle of the cannon is, and this would vary the velocities. Say a 45 degree angle, then cy=cx, so we get (from 5=cy*cx) cx=sqrt(5), then the velocity is sqrt(10) (from sqrt(cy^2 + cx^2)).

Does that make any sense at all?
firemeboy
Posted: Tue Sep 05, 2000 2:51 pm    Post subject: -14

Mercuria, I am flattered that you are talking to me as if I were intelligent. However, when it comes to math, I lack severely and have to be shown how it works.

You have a distance of say 100 meters, and a 'cannonball' of say 5 kg. What would be the velocity you would need to hit the target. Assuming no wind.

Mercuria
Posted: Sat Sep 02, 2000 2:06 am    Post subject: -15

oh, right... the acceleration is the acceleration of gravity (-9.80 m/s^2 or -16.2 ft/s^2)

i'm not sure of the ft/s^2 one, but i know the metric one for sure ;->

x = displacement... so there's positive and negative...
Mercuria
Posted: Sat Sep 02, 2000 2:03 am    Post subject: -16

they're pretty simple. you have a vertical component (x = vt + .5at^2) and a horizontal component (x = vt). (v = velocity)

weight makes no difference if you use velocity and angles.

basically, you use the pythagorean theorem to figure out the vertical and horizontal components...

have fun ;->
firemeboy
Posted: Fri Sep 01, 2000 8:55 pm    Post subject: -17

It is interesting how Baz said that this forum dosen't get used enough. I have been crafting a post for this forum all week.

I am an Instructional Designer for a large retail company. I design computer guided training and it is boring as you may imagine (both the job, and the training). Every night I go home and look at my wish list of games to buy and think why cant education be fun? Not fun as in, "Let's add candy bars together!" but fun as a boticelli is, or some of the other riddles here. Can anybody who has been posting here for more than a week honestly say they have learned nothing?

So here is my idea. I have been attending a conference and had an epiphany. Why not make learning nothing more than solving engadging problems (I teach customer service, not spelling).

I am thinking about going back to school and getting my Phd. I thought that maybe for an interesting project that would kick off the whole process would be design a game that teaches some basic physics.

Did anybody use to play the old computer game where you had two cannons on the screen, and you tried to blow up the other guy before he got you? A wonderful game, that was engaging, but it didn't teach you anything. I am going to design a game that allows you to play the game at different levels of complexity while teaching you some basic physics formulas.

So I ask two questions of my fellow GL members. First what do you think? Good idea? Bad idea? Would you people be interested in trying out some of the prototypes?

Second... I suck at math. What are some of the formulas I am looking at? The input (At least for the first levels) would be nothing more than angle of cannon, velocity of cannon. The desired outcome would be the distance of the shot. Later I would have the student account for wind that blew both from the left and right, and from the front and back. Also, I may change the weight of the cannon balls, I still can't be sure. Are we looking at some extreme formulas, or just some pretty basic ones. Any help would be appreciated.

fire