Horizontal Bar: Vertical Velocity

[animater31405]

I see this whole thing on, how are they really going to know if somethings going too fast if the object isn't going 20 ft (sarcastically) way up in the air and hitting someone in the head. I think the judges will be very lenient on this role and cut alot of slack within safety reasoning.

[lips]

Just curious, I know that the vmax=10ft/sec, but does that apply to non-projectiles aswell? for intance, if we were to use a tube-in-tube to raise a hook up to grapple the bar, but the hook was (on the first trip) attached to the hard tubing, which was in turn attached to the roboti, would it be considered a projectile? if so, we were be running it with a motor and therefore still running at a constant 10ft/sec, so it would need to take 1 sec to get up there. if not, could we raise it as fast as possible?

[animater31405]

I don't think that would be considered a projectile by any means but that's just me.

[tkwetzel]

The rule states that all mechanisms must not exceed 10 f/s. That would include any telscoping arms, projectiles, or anything else.

[Yan Wang]

Originally Posted by Mike Rush

Use the energy equations

mgh = 1/2 (mv^2)
gh=(v^2)/2
h=v^2/2g
h=100/64.4
h=1.55 ft

if the object goes up over 1.55 feet it is travelingmore than 10 ft/s

(I hope this is right... It's been a long time )

Er... I am currently taking AP Physics so I am not very experienced with it. But it seems to me that you are saying: Potential Energy = Kinetic Energy... Well, when the robot is hanging it's got PE, but no KE... I think it was lucky that when solving for 'h', you came out with v^2/2g

I used Andy's method of applying the equation "V^2=Vo^2+2aX" and derived the same formula.. Hmmmmm.

[Nick R.]

to :Mike Betts
from: Team #0177 (Bobcat Robotics)

You do have a fair point about it being dangerous and all but than in the simulations they did 'live' from new hamhire or whereever it was; the person that hung on to the bar actualy threw the strap he than secured toh is little bike and hung on the bar by it.... and i think that was woody that said "I think mike there didnt go over 10 ft/s when he threw it" so i'm pretty sure its going to be alowed one way or another, we just have to watch fo more safety rules and stuff like that to observe when we're building it

[Joe Ross]

Originally Posted by monsieurcoffee

Er... I am currently taking AP Physics so I am not very experienced with it. But it seems to me that you are saying: Potential Energy = Kinetic Energy... Well, when the robot is hanging it's got PE, but no KE... I think it was lucky that when solving for 'h', you came out with v^2/2g

I used Andy's method of applying the equation "V^2=Vo^2+2aX" and derived the same formula.. Hmmmmm.

Mike's equation comes from the fact that energy must be conserved. If you take your origin to be the height at which the projectile is launched, then the projectile at the beginning would have no potential energy but would have kinetic energy. At the peak, it wouldn't be moving and so wouldn't have kinetic energy but would have potential energy. Since energy must be conserved, those are equal.

Remember, in physics, there are often many ways to find the same answer. Your way is right also, but harder (to me).

[andy]

Originally Posted by monsieurcoffee

Er... I am currently taking AP Physics so I am not very experienced with it. But it seems to me that you are saying: Potential Energy = Kinetic Energy... Well, when the robot is hanging it's got PE, but no KE... I think it was lucky that when solving for 'h', you came out with v^2/2g

I used Andy's method of applying the equation "V^2=Vo^2+2aX" and derived the same formula.. Hmmmmm.

He is saying that potiential energy equals kenetic energy!
Conservation of energy vs. kinematics.
This are just two different ways to solve the same problem.
We are both correct except I am more right because I used the metric system.

Just kidding, seriously though, the metric system is so much easier to use then the imperial system. There is nothing easier then multiplying/dividing powers of ten then fractions (shudder).

Useless info:
-Americans are better at fractions then Europeans because Americans have to be fluent with fractions because of their/our measurement system. Interesting
-Our robot will probably be in inches because of the availability of metric vs. imperial fasteners, drive componants, extruded aluminum, 80/20 etc.

Good luck this year!
-Andy

[ahecht]

Originally Posted by lips

Just curious, I know that the vmax=10ft/sec, but does that apply to non-projectiles aswell? for intance, if we were to use a tube-in-tube to raise a hook up to grapple the bar, but the hook was (on the first trip) attached to the hard tubing, which was in turn attached to the roboti, would it be considered a projectile? if so, we were be running it with a motor and therefore still running at a constant 10ft/sec, so it would need to take 1 sec to get up there. if not, could we raise it as fast as possible?

I hope so. I was nearly impaled in the pits at Nats by a robot with an arm that extended at atleast 10fl/sec. It can be even more dangerous than a projectile, sin

[Yan Wang]

Originally Posted by andy

Just kidding, seriously though, the metric system is so much easier to use then the imperial system. There is nothing easier then multiplying/dividing powers of ten then fractions (shudder).

I agree... when everything else is base 10, the american system is just confusing. I came out with 15.5 feet the first time I calculated it due to conversion errors. Lol. Since FIRST promotes culture change, why not change the FIRST standard system of measurement to metric?? It wouldn't hurt the Canadian, British, or Brazilian teams at all

[michael_obrien]

mgh = 1/2 (mv^2)
gh=(v^2)/2
h=v^2/2g
h=100/64.4
h=1.55 ft

if the object goes up over 1.55 feet it is travelingmore than 10 ft/s

(I hope this is right... It's been a long time )

i think this is wrong...

first convert feet/sec to meters/sec then do the equations then convert back, you get about 15 feet high if i'm right.

you have to do this because v is squared (if i'm right)

but i'm probably wrong.

[Yan Wang]

Originally Posted by michael_obrien

i think this is wrong...

first convert feet/sec to meters/sec then do the equations then convert back, you get about 15 feet high if i'm right.

you have to do this because v is squared (if i'm right)

but i'm probably wrong.

No, the equation is right.

Vmax = 10ft/s
g = 9.8m/s^2 = 32.15ft/s^2

You get 1.55 ft (100/63.3)

[michael_obrien]

oh, right, that was dumb, i didn't notice that it'd been corrected originally