Joined: 18 Jun 2004
Posts: 3842
Location: Queensland near Brisbane
hair-splitting
I have been called pedantic before. This thread is all about those little inconsequential things that press our buttons.
The simbol for kilo is a lower case k. The upper case K is the simbol for absolute temperature. Grams is lower case g. Upper case G is giga. Lower case m is meters. Upper case M is mega.
Thu Jul 08, 2004 1:52 pm
Spockie-Tech Site Admin
Joined: 31 May 2004
Posts: 3160
Location: Melbourne, Australia
There you go, I didnt know about that.
it seems a little inconsistent though.. low case k for kilo, uppercase M for mega ? I know the french are supposedly the originators of the SI units system, but that seems stupid to me.
the apostrophe in the "PHOTO'S" link on the top of the forums _________________ The Engine Whisperer
- fixer of things
Thu Jul 08, 2004 2:43 pm
colin
Joined: 16 Jun 2004
Posts: 102
I though G was for gravitaional constant. I guess no one will mind if we ditch that though
Thu Jul 08, 2004 7:07 pm
Spockie-Tech Site Admin
Joined: 31 May 2004
Posts: 3160
Location: Melbourne, Australia
Oh I dont know.. Knowing the gravitational constant could be very useful for Launcher style robots.. you could calculate the effective kilojoules of impact when 12kg hits the ground after accelerating under 1g for X meters
Thu Jul 08, 2004 11:15 pm
Ajax Experienced Roboteer
Joined: 17 Jun 2004
Posts: 298
Location: Sydney
Ok, lets say a 12kg robot is thrown 3 meters in to the air, gravity will still set the terminal velocity, 9.8 m/sec. now the acceleration of the bot returning to earth will be 16.01 m/sec and the force of the bot hitting the ground would be 192.08 Newtons.
I did make the assuption that the floor would not move and the bot came to a stop in mid air.
I believe I calculated it correctly. _________________ It's all about the destruction.
Fri Jul 09, 2004 2:05 am
mytqik
Joined: 26 Jun 2004
Posts: 127
Ajax,
I come up with a different answer, but mine seems rather large:
Vf^2 - Vi^2 = 2 x A x D
Vf^2 - 0 = 2 x 9.81 x 3
Vf^2 = 58.86
Vf = 7.672m/s
Double Check:
d = Vi x t + 0.5 x A x T^2
3 = 0 + 0.5 x 9.81 x T^2
T = 0.78 sec
Vf = Vi + at
Vf = 0 + 9.81 x 0.78
Vf = 7.65m/s
F x D = - 0.5 x M x V^2 (where d is the distance into the floor it travels or the amount the frame crumples)
F x 0.05 = - 0.5 x 12 x 7.67^2
F = 7059kg
This is where I go wrong I think. Surely a 12kg object falling from 3m does not exert a 7 ton force. I based this on the conservation of energy principle.
At 6 o'clock this morning I woke up and and remembered that I did the calc incorectly. I used 9.8m/s as the velocity.
It should of been the acceleration.
Asumptions
the ground doesn't move.
there is no frame crumple
the correct formula is
a = verlocity
v = acceleration (9.8m/s/s)
d = distance (3 m)
f = force (Nowtons)
m = mass (12 kg)
f = m x (v / (sqrt(0.5 x v) / d))
f = 12 x (9.8 / (sqrt(0.5 x 9.8 ) / 3))
f = 92.02 Newtons
Now I hope I have it correct.
The D is causing the problem. I get the same when I calc the bounce and crumple. _________________ It's all about the destruction.
Fri Jul 09, 2004 11:29 am
mytqik
Joined: 26 Jun 2004
Posts: 127
Ajax,
Silly question, but aren't you using the velocity at impact (m/s) as opposed to acceleration (m/s/s) for the force calculation? If so doesn't that mean that the sum you just did calculated the momentum not force? Because your units are kg.m/s (12kg x 7.87m/s)??
The reason your calcs are the same for bounce & crumple is that bounce assumes no crumple & all force exerted by robot into the floor is returned to the robot. This will never happen. Eg a purely elastic collision. You also stated that you are assuming no crumple which makes the calc the same as the bounce. If you assume no bouce & no crumple you are dividing by 0 in the momentum equation which you can't do.
Gee i wished I payed more attention in all those dry & boring kinetics classes.
Fri Jul 09, 2004 12:14 pm
kkeerroo Experienced Roboteer
Joined: 17 Jun 2004
Posts: 1459
Location: Brisbane
Force? In an question about impact? No, no, no.
When something hits the floor that'll be a huge amout of deaccleration. The speed for a robot falling 3m would be about 7.67m/s. Then stopping that in 0.1 seconds (slow impact there) would cause about -76.7m/s^2. Thats a force of 920N. This is all good, but it makes no sence. How do you know how long it'll take the robot to stop? What if its 0.01 seconds? Thats 9200N or about 1 ton of force.
No, impacts are measured in energy. And the amount of energy in a falling robot is simple.
Mass x Height x Gavitational Constant (for sea level anyway)
12kg x 3m x 9.81m/s^2
24.81 Joules.
I believe that. Dropping my robot from the roof wont do more damage then Nick's spinner and it about 273 Joules (I think). Drop it from 6 stories and then the wheel would shatter and the gearbox be ripped open. _________________ Get Some!!!
Secretary of the Queensland Robotics Sports Club inc.
Fri Jul 09, 2004 1:28 pm
Ajax Experienced Roboteer
Joined: 17 Jun 2004
Posts: 298
Location: Sydney
kkeerroo,
I have been looking into doing some basic calcs.
As I have done any of this sort of calc for over 15 years so it is very possible that some thing is wrong.
and I was not worrying about things like resistance, dimensions of bot, material types, design, etc which affect the end result. I'll leave that up to the mathematicians.
My be I should just leave all the maths and do the suck it and see approach.
if the bot hold together, after a battle it must be strong enough.
mytqik,
I found on the web (it was a maths site.) a formula 'f = ma' and the site said it was Nowtons.
If it is wrong I'm sorry _________________ It's all about the destruction.
Fri Jul 09, 2004 1:50 pm
DumHed Experienced Roboteer
Joined: 29 Jun 2004
Posts: 1219
Location: Sydney
well really it's all inter related, so you can work it out using whichever units you prefer.
An impact will exert a force, which is what does the damage, but it's usually calculated as energy because the actual force is determined by how fast the deceleration in the impact is - and that's hard to predict.
The problem is that impact energy is also a useless thing to know, because all objects of the same mass falling the same distance will have the same energy.
Their design determines how much force is created, and how much force it can take without damage.
In the end the only way is to build it and see _________________ The Engine Whisperer
- fixer of things
Fri Jul 09, 2004 2:32 pm
3Faze
Joined: 26 Jun 2004
Posts: 99
Location: Lincolnshire, UK
quote:Originally posted by DumHed:
the apostrophe in the "PHOTO'S" link on the top of the forums
That's justifiable, the apostrophe can be to denote the ommission of "graph" from the word
Fri Jul 09, 2004 9:39 pm
mytqik
Joined: 26 Jun 2004
Posts: 127
KKeerroo, isn't the resultant force of the impact the MOST important thing here? If you build a completely rigid frame so that no crumple/deformation is allowed then it would need to withstand much higher loadings than a flame that is allowed to flex & absorb the impact?
That is the principle behind crumple zones in cars. Allow the monoque of the car to absorb the impact, thus decreasing the amount of force exerted to the passengers during an impact.
The energy part of the equation is only of use if you can calculate the energy absorbed by the frame of the robot during deformation. The more the frame distorts the more energy is aborbed. If the frame distorts more than its elestic limit then the frame will bend, if not it will flex back to its original shape. This series of equations is much more complicated than I am prepared to do. As it is not a linear equation & it has to be recalculated for every shape the distorted chassis takes.
Just my 2 cents worth.
Fri Jul 09, 2004 11:33 pm
kkeerroo Experienced Roboteer
Joined: 17 Jun 2004
Posts: 1459
Location: Brisbane
Yeah, I see where you coming from, but I think your working backwards to the way I see it.
Mr Newton says the energy is conserved. Force isn't. When a car hits a wall you get both the force and energy of the impact distributed into the car. The frame gets a certain force into it which applies a stress into is parts. If the stress is higher then what that part can take then it begins to stain (bend or stretch) which aborsbs the energy. When the energy lowers so does the force since the force and energy are linked together.
Force x Displacment = Energy
Since they are both linked you don't need to work out both. And judging from all the different ansrews above with the force calcs it would be easier to calculate the energy that is applied to the system, which for a falling robot is simple. Since energy is conserved the energy of the robot as is hits the ground is the same as it the energy at the top of its flying arc. That energy is gravitational protentional energy which is, as is said earlier,
Mass (kg) x Hight (m) x 9.81 (m/s^2)
If you started with force from the start you need to know all sorts of things like deacceleration as it hits the ground, which for a theoretical calc is damn hard to do. _________________ Get Some!!!
Secretary of the Queensland Robotics Sports Club inc.
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