4 newtons of force???

[kerrigan93]

The rules say that the mini bot has to push the sensor with a force of 2-4 newtons. I think that it would be a good idea to try to go for a force of 4 newtons but I am curious as to how much force 4 newtons is.

What are some examples of how much force this is and what mass and velocity a bot would have to be moving to create such a force? Is this accomplishable by the "spring loaded" bots that people are talking about?

[KrazyCarl92]

.899 lbs of force, idk specific examples

[Cyberphil]

4 Newtons is about .9 of a lb. So a fairly strong force for that small of a robot, but I am sure you can get that much of a force.

[big1boom]

What are some examples of how much force this is and what mass and velocity a bot would have to be moving to create such a force? Is this accomplishable by the "spring loaded" bots that people are talking about?

http://upload.wikimedia.org/math/a/2/8/a28d8d4e4a422c639a1d24d9ce9e3d8e.png
Otherwise known as F=ma

If you have a mass of 2kg, then this means you need an acceleration of 2m/s^2

Now, you assume that you are working with a free floating object (no friction/gravity), and approximately 1/4" or 0.0063m of stopping distance (based on the rules).

This means, that you need to solve for u, or initial velocity in this formula.

v=final velocity=0
u=initial velocity=unknown
a=acceleration=2
s=distance traveled=0.0063

v^2 = u^2 + 2as

Based on that I am coming up with about 0.16m/s
Now, this is actually low, due to the fact that the calculation I just did is ignoring gravity.

I am using a formula from
http://tutor4physics.com/motion1d.htm
So, I dunno if it is correct, but it sounds about right. Admittedly, I need to brush up on Newtonian physics

[Matt H.]

Physics Content:
1 Newton=0.225 lbs
As such, one way to trigger the plates is to push upwards with ~.9 lbs of force.

However, the questions you've asked especially about moving robots requires some additional discussion.
A robot moving upwards has some momentum (P) defined as the robot's mass multiplied by its velocity:
P=m*v
Assume that when the robot hits the button it exerts a constant force F until coming to a stop. Using Newton's second law:
F*delta t=P where delta t is the time required for the robot to come to a complete stop (v=0, P=0).
or
F=P/(delta t)

As such, a robot moving fast (high P) could trigger the button even if it could not produce the necessary force when stationary.

[kerrigan93]

so basically we have found that it does not take a lot of speed to generate 4 newtons of force with a 4.5lb mini bot?


This is a lot of science content and I don't think that i comprehend all of it seeing that I am still in high school and i am currently in week 2 of my physics class

[big1boom]

What I showed you is one dimensional Newtonian mechanics. You will learn it in your physics course. (probably)

[marwahaha]

i will try to make sense.
notes:
110" = 2.794meters
1/4" = .00635 meters
g=9.8 is the acceleration of gravity
m=mass, v=velocity, h=height from base
Fd=work done (force x distance)
(i.e. 4Newtons at top of the pole x .00635 meters to move plate)

Energy is always conserved. E = E
0.5mv^2 + mgh + Fd = 0.5mv^2 + mgh + Fd
(at bottom of the pole) = (at top of the pole)

0.5mv^2 + 0 (h=0) + 0 (no work done)=
0 (v=0) + m(9.8)(2.794meters) + (4Newtons)(.00635meters)

.5mv^2 = 27.3812m + .0254

mv^2 = 54.7624m + .0508

v^2 = 54.7624 + (.0508/m)

v = sqrt(54.7624 + (.0508/m))

this looks complicated: however, m (mass) is in kilograms.
Unless your bot is less than .1kg = .22lbs, that number will be negligible.

v approximates sqrt(54.7624) =7.4.... m/s, i'm going to round up to 7.5 to be safe.

you need MINIMUM 7.5 m/s = 25 fps upward launching velocity if you were to launch an object from the base that does not touch the pole but effectively trips the sensor. (like a cylinder ring) This accounts for gravity not for air resistance (low) and friction (rubbing against the pole), if you had friction you should focus on programming a robot that actually climbs rather than launches.


concerning a push from a moving/climbing/(or launched) robot,
someone may have already answered this.

the energy transferred to the sensor is = Kinetic Energy (due to motion)
E = 1/2 m v^2 = Fd = (4N)(.00635meters) =.0254 Joules

So assuming your robot weighs about 2 pounds (1 kg) then
1/2 (1) v^2 = .0254
v^2 = .0508
v= .2253 m/s = 0.75 fps

The more the robot weighs, the less the required velocity.

In conclusion:
If you can get a robot moving and climbing the tower high enough in 10 seconds, you probably WILL trip the sensor.
If you launch something (unclear, GDC will clear it up soon) then you need about 25 fps from the base for that object to trip the sensor.

Good luck.

[Hawiian Cadder]

has anyone thought about breaking the top of the pole, i know if the mini-bot setup my team has envisioned as of now is made reality there will be a LOT more than 4 newtons of force.