[Torg] Jumping in Torg
Benjmain Grant
benn at 4efix.com
Wed Dec 10 15:44:14 EST 2008
Very thought-provoking! Here are some of those thoughts:
To best add to the line of discussion, I need to somewhat recreate to best
understand it. So here goes.
Imagine a ramp:
...
.........
...............
.....................
...........................
If you leave the top of the ramp doing speed "S", then we can imagine a
right triangle that looks a lot like the ramp where the speed "S" is the
hypotenuse (the long side), the vertical side on the right represents the
vertical component of the speed and the horizontal side at the bottom of the
triangle represents the horizontal component of the speed. Let's call the
horizontal component of the speed "X" and the vertical component of the
speed "Y".
If we know the angle of the ramp is N, then basic trig from high school
tells us that:
Sin(N) = (length opposite the angle) / (hypotenuse) = (vertical component) /
(speed leaving the ramp) = Y/S. And if Sin(N) = Y/S, then S*Sin(N)=Y.
Similarly,
Cos(N) = (length adjacent to the angle) / (hypotenuse) = (horizontal
component) / (speed leaving the ramp) = X/S. Cos(N) = Y/S means that
S*Cos(N)=X
We know that the time spent in the air (the "hang time") multiplied by the
horizontal speed component (X) will give us our total jump distance. How do
we calculate the hang time?
The time it takes a falling object to reach a speed is the same time it take
an object traveling against gravity to reach its max height. How long does
it take an falling object to reach a speed of Y (the vertical speed
component)? The earth acts with an acceleration of about 10 m/s per second.
Let's call this value of the earth's attraction "A". Then a simple division
of Y/A tell us exactly how long it takes a falling object to reach the speed
of Y, and by extension, also tells us exactly how much times passes between
launching an object upward, and that object reaching is max height before it
begins to fall down again.
However, the hang time we are looking for is twice that because it is the
time for the object to get to it's max height and then to fall back down.
So hang time = 2Y/A.
(Note: technically hang time is 2Y/A *only* if one is jumping from the top
of one ramp to the top of an equal height ramp on the other side. If there
is no equal height ramp on the other side, you gain a slight amount of extra
"hang time" before hitting the floor equal to the height of the ramp divided
by A. So if the ramp is 10 meters off the ground and you are landing on the
ground, you will get a slight amount extra time, which translates into a
slight amount further distance. However, that gets *much* more complex to
calculate - as in using the quadratic formula and more. A simpler method of
determining this extra distance can be used after answering the original
question, and will be addressed later.)
All we have to do is multiply hang time by the horizontal speed component
"X" to get the max jump distance: X*2Y/A, or to simplify, 2XY/A.
However, we do not know X and Y - although we do know that Y=S*Sin(N) and
X=S*Cos(N), so we can substitute those value into our 2XY/A formula:
2 * (S*Cos(N)) * (S*Sin(N)) / A
Rearranging slightly, we get:
2 * S * S * Sin(N) * Cos(N) / A
Or...
2S^2(Sin(N))(Cos(N))/A
Since we know A = 10 m/s^2...
2S^2(Sin(N))(Cos(N))/10
Or...
S^2(Sin(N))(Cos(N))/5 ... which is the same as .2*(S^2)*(Sin(N))*(Cos(N))
One interesting question is what angle of ramp gives you the longest
possible jump? Another way to ask this is to ask for what value of N is
Sin(N) times Cos(N) the largest possible value? The answer happens to be 45
degrees, where Sin(N)Cos(N) = .5
Plugging this into to our equation yields:
.2*(S^2)*.5 = .2*.5*(S^2) = .1(S^2)
So, if the object leaves a 45 degree ramp at a speed (in m/s) of S, to
determine his total jump distance square his speed and multiple by .1 (or
divide by ten, same thing.) So leaving a 45 degree ramp at 100 mph (about
45 m/s) should yield about a 45^2/10 distance = 202.5 meters, about 665 feet
- not including air resistance.
A simple table of the Sin(N)Cos(N) factor (call it the SCN factor) could be
useful, here is one:
RAMP INCLINE SCN
5 or 85 degrees use 0.09
10 or 80 degrees use 0.17
15 or 75 degrees use 0.25
20 or 70 degrees use 0.32
25 or 65 degrees use 0.38
30 or 60 degrees use 0.43
35 or 55 degrees use 0.47
40 or 50 degrees use 0.49
45 degrees use 0.5
Simply look up the degree of the ramp or incline above to plug in to the
formula:
(.2) times (Speed in m/s, squared) times (SCN factor from table) = total
horizontal distance
And that will give you your max jump distance with no need to do trig at
all. Easy peasie.
To get max height, for jumping over obstacles and whatnot, it's a little
different. The formula d=1/2at^2 tell us the distance travelled by an
object under constant acceleration for a time of "t". In this case, we are
experiencing constant deceleration do to gravity.
We start by determining at what point in time is our max height happening.
We already figured that out above - it's the vertical speed component Y
divided by the deceleration due to gravity A. Time to max height = Y/A.
If we were to drop an object and measure at what distance of fall it is at
when it reaches speed Y, that will be the same distance it take for an
object of upward speed Y to reach it's max height before it begins to fall
back down. D=1/2AT^2. In this case, we know that the time taken is Y/A.
Therefore:
D = 1/2 * A * (Y/A)^2 = 1/2 * A * Y^2/A^2 = 1/2 * Y^2/A = (Y^2)/2A. This is
the max height gained from the top of the ramp. To find the height from the
ground of course you will have to add the height of the ramp itself (H):
H + ((Y^2)/2A) = the max height from the ground.
We already know the max length of the jump from the top of one ramp to the
top of another of the same height is 2XY/A. What if there is no second ramp
and the landing is on the ground itself?
We know the time to max height is Y/A, but what is the time is take to fall
from that max height not to the level of the ramp where we started, but to
fall to the ground itself? Let's call the falling time "F". Then the
standard d=(1/2)AT^2 formula can be rewritten as (total height) = (1/2)AF^2
Since we know the total height of the fall is H + ((Y^2)/2A) we can sub that
in to know the F, the time of the fall:
H + ((Y^2)/2A) = (1/2)AF^2
Solving for f yields:
2H + (Y^2)/A = AF^2
2H/A + (Y^2)/(A^2) = F^2
2HA/(A^2) + (Y^2)/(A^2) = F^2
(2HA + Y^2)/(A^2) = F^2
[(2HA + Y^2)/A^2]^.5 = F
Now that we know both T, the time till max height from ground (starting at
top of ramp) and F, the time for the object to fall from the max height all
the way to the ground, we can add those together to find the total time in
the air of the object:
F = [(2HA + Y^2)/A^2]^.5
T = Y/A
T+F = [(2HA + Y^2)/A^2]^.5 + Y/A
This total time in the air multiplied by the horizontal speed component
yields the true total horizontal jump distance when not landing at the same
level that one took off from:
X * ([(2HA + Y^2)/A^2]^.5 + Y/A)
X[(2HA + Y^2)/A^2]^.5 + XY/A
Finally, to sub in the true value for X and Y, as well as replace A with a
metric value of 10 m/s^2:
Y=S*Sin(N) and X=S*Cos(N)
S*Cos(N)* [(2H(10) + [S*Sin(N)]^2)/10^2]^.5 + [S^2*Sin(N)*Cos(N)]/10
Or...
S*Cos(N) * [(20H + [S*Sin(N)]^2)/100]^.5 + [S^2*Sin(N)*Cos(N)]/10
Or...
S*Cos(N) * [ [(20H + [S*Sin(N)]^2)/100]^.5 + S*Sin(N)/10]
Or...
S*Cos(N) * [ [20H + [S*Sin(N)]^2]^.5/10 + S*Sin(N)/10 ]
Or...
S*Cos(N) * (([20H + [S*Sin(N)]^2]^.5 + S*Sin(N)) /10 )
Or...
0.1 * S*Cos(N) * [ [20H+([S*Sin(N)]^2)]^.5 + S*Sin(N) ]
*THAT* is, finally, the distance formula for the total horizontal distance
given a launch speed of S, a ramp incline of N degrees, and a ramp height of
H, given that you will be landing on the ground and not a second ramp at the
same height.
So now we have three answers:
A) to find the total distance assuming that you land at the *same* level you
took off from:
RAMP INCLINE SCN
5 or 85 degrees use 0.09
10 or 80 degrees use 0.17
15 or 75 degrees use 0.25
20 or 70 degrees use 0.32
25 or 65 degrees use 0.38
30 or 60 degrees use 0.43
35 or 55 degrees use 0.47
40 or 50 degrees use 0.49
45 degrees use 0.5
Simply look up the degree of the ramp or incline above to plug in to the
formula:
(.2) times (Speed in m/s, squared) times (SCN factor from table) = total
horizontal distance
B) To find the total max height of the jump, use the quick cheat sheet
below:
RAMP INCLINE RIF (Ramp Incline Factor = Sin(N))
5 degrees use 0.09
10 degrees use 0.17
15 degrees use 0.26
20 degrees use 0.34
25 degrees use 0.42
30 degrees use 0.50
35 degrees use 0.57
40 degrees use 0.64
45 degrees use 0.71
50 degrees use 0.77
55 degrees use 0.82
60 degrees use 0.87
65 degrees use 0.91
70 degrees use 0.94
75 degrees use 0.97
80 degrees use 0.98
85+degrees use 1.00
... and use the formula: [(S*RIF)^2]/2 to get max jump height from liftoff
point.
Make sure that S, your ramp speed, is in meters/second. If you are using a
ramp, add H, the ramp height to get the total height of jump.
C) And finally, to find the total complete distance jumped when taking off
at a higher level than you land, including the small bit extra, use:
0.1 * S*Cos(N) * [ [20H+([S*Sin(N)]^2)]^.5 + S*Sin(N) ]
Where S = ramp speed in m/s, H = ramp height in meters, N = ramp incline -
sorry, there is no cheat sheet for this one.
Enjoy.
-Benn Grant
eFix Computer Consulting
benn at 4eFix.com
603.283.6601
-----Original Message-----
From: torg-bounces at justintimeadventures.com
[mailto:torg-bounces at justintimeadventures.com] On Behalf Of Travis James
Hall
Sent: Wednesday, December 10, 2008 10:18 AM
To: torg at justintimeadventures.com
Subject: RE: [Torg] Jumping in Torg
> -----Original Message-----
> From: torg-bounces at justintimeadventures.com
> [mailto:torg-bounces at justintimeadventures.com] On Behalf Of
> Sam Frazier II
> Sent: Thursday, 11 December 2008 1:11 AM
> To: torg at justintimeadventures.com
> Subject: Re: [Torg] Jumping in Torg
>
> I would agree that a velocity and strength would be needed to
> determine the distance in a Long jump. There should be a
> formula for it though. *ponder* Off the top of my head, Value
> of speed + Value of Str - Value of Weight = Distance
> traveled? Which would be V * Str / effects of gravity.
> *thinks* something like that.
>
> I haven't done the math, but I would start with something like this.
>
> I think there should be a fairly simple way to guestimate
> without getting sin/cos/tan involved. *winks at Travis or
> whomever mentioned sin/cos/tan in a formula. I forget* though
> your forumla seems correct to me, I'll agree with your point.
Well, I did point out that sine maxes out at 1 for a 45deg jump, so you can
just ignore that factor to get your max and eyeball for a less than optimal
angle.
But now that you mention it, the reason we have trig functions involved is
because we need to split the velocity into horizontal and vertical
components, and sine and cosine represent those relationships. If we have
another means of getting values for the horizontal and vertical components
of velocity, we can avoid those, and since it's hard to judge the angle of a
long-jumper's leap, that may be an easier approach anyway.
An alternate formula for working up projectile range is d = -2v(y)v(x)/a
where v(y) and v(x) are the vertical and horizontal components of velocity
respectively. (Dratted ASCII, no subscripts - sorry about the
difficult-to-read notation.)
We can reasonably suppose that v(x) is equivalent to running speed.
(Probably a touch less, as the jumper would tend to lose a little horizontal
velocity in that last step/leap, when effort is devoted to pushing up as
well as maintaining speed, but close enough in terms of Torg scales.)
v(y) could reasonably be supposed to be proportional to Strength, as leg
strength would be used to provide that initial upward force. Whether it is
reasonable to simply reason Strength through the value chart, whether a
limit value should be applied, whether a modifier should be applied, or
combinations, is up to the GM (because this is well into house rule
territory). You'd have to try some examples and see if the figures come out
about right. (The world record for long jump is a little under 9m. If
characters are regularly beating that, you've got a problem.)
a is (as previously noted) about -10m/s, which means our long jump distance
is found with v(x)v(y)/5 . But remember, that's working with measures (in
Torg terms). The multiplication becomes nice, neat addition when working
with Torg values. The equivalent in Torg values would be something along the
lines of (DEX-5)+(STR-?)-4, with a limit value on the Dex of 10 (that's the
running limit value) and the -5 being because Torg running speed is in
metres per 10-second round and we need m/s. The question mark represents an
unknown factor for unit conversion for Strength into upward motion, similar
to the 5 for converting running speed to m/s. If we suppose Strength
converts to speed similarly to Dex, that becomes (DEX-5)+(STR-5)-4 =
DEX+STR-14.
A little suggested calibration... That gives a max value (for normal humans)
of 10+13-14=9, which becomes a jump distance of about 60 meters - obviously
too high. Before pushing, we need this to max out 3 to get long-jump results
in the ballpark. That's lower by six, so apply a limit value to the Strength
of 13-6=7.
So what you get is, add Strength (limit 7) plus Dex (limit 10) and subtract
14 (unit conversion and accounting for gravity), and read through the Torg
value chart to translate to an actual distance. Pushing should be done on
the Speed column using Strength (because that's the lower limit value, and
it would only be fair to use the more limited stat). Without pushing
characters would jump around 4 metres, with pushing up to around 10 metres,
which is about as close as we can expect to get to anything when reading
through the Torg value chart.
Or you can fiddle around with alternate factors and limit values. There's
other ways to get similar basic results (and each will give a different
effect from pushing).
But that's not so helpful for jumping cars off ramps, because cars don't
have the applicable jumping ability to allow Strength to add to the height
of the jump. For cars, it is back to the trig (if our available information
is just angle of the ramp and speed coming off the ramp).
Travis Hall
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