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Formula to find horsepower hours

02/06/2009 6:55 PM

My high school battlebot team is using a spinning weapon on our robot this year. I would like to know how to calculate the horsepower-hours or watt-hours used when I accelerate a cylinder around its center with a known MOI from 0 to 2,500 rpm. Thanks!

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#1

Re: Formula to find horsepower hours

02/06/2009 8:40 PM

This sounds like a legitimate homework-related question because the numerical answer isn't being sought; only the method to get there.

First, when you are asking for hp-hours, or Watt-hours, recognize that you are asking for an answer in terms of energy. Both hp and Watts are units of power, and power multiplied by time equals energy.

The energy equation for this problem is analogous to, but not quite the same, as if you were asking how much energy it takes to accelerate a stationary item to some final velocity. In that case, the answer would be (ignoring all sources of friction):

E = (1/2) m * v^2

the familiar expression for the kinetic energy (E, in Joules) of a moving mass (m, in kilograms) at a velocity (v, in meters per second). (All units in the MKS or SI system.)

When you have a stationary rotating mass with a moment of inertia, the equation for its energy is

E = (1/2) I * w^2

where

E is energy as before,

I is the moment of inertia, units of mass * length^2 or kg-m^2 in the MKS system, and

w is my poor substitute for the lower case Greek letter omega, which is used to denote angular velocity, with units of radians per second (how fast the unit is spinning).

Note the units work out as they must with E in Joules, because the moment of inertia "I" is mass times length squared, whereas the angular velocity has units essentially of "per unit time" so that when you multiply the units on the right-hand side of the equation, you get the same as when multiplying the units on the right-hand side of the familiar kinetic energy equation.

So that's pretty much it. With the weapon at rest, its kinetic energy is zero, and you can now calculate the kinetic energy with it spinning, and the difference is the amount of energy it took to get the weapon spinning (again ignoring friction, which includes air resistance, which is likely to be the dominant source of lost energy if you have decent bearings).

You have to be careful with units. 2500 rpm is a measure of angular velocity, but you will have to convert to radians per second. I'm going to let you do that, giving just the key conversion factor. There are 2*pi radians per revolution, by definition of what radian angle measure is.

So I believe you now have all the info you need. If you don't, come back with a question, but remember that CR4 is not about getting homework done, but about learning.

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#2

Re: Formula to find horsepower hours

02/06/2009 9:30 PM

The rotational kinetic energy of a spinning object (of any shape but having a known MOI) is given by:

Ek = ½ * MOI * ω2 , where (Eqn 1)

Ek is the kinetic energy in Joules,

MOI = the moment of inertia

ω = angular velocity in radians per second

---

But first, let's do some conversions:

We want our rotational energy expressed in watt-hours, not Joules, and so we need to convert from Joules to watt-hours:

1 Joule = 1 watt-second, therefore, 1 watt-hour = 3600 watt-seconds = 3600 Joules

And so, J Joules = J watt-seconds = (J/3600) watt-hours (WH). (Eqn 2)

---

We also have RPM, but we need to express this in terms of radians per second, and so:

RPM (rev/min) x (1 min/60 sec) x (360°/rev) x (2π rad/360°)

= RPM x 2π rad/60 sec

= RPM x π/30 radians per second (Eqn 3)

---

We now combine these with Eqn 1:

Ek = [½ * MOI * (RPM x π/30)2] / 3600

= MOI * RPM2 * π2/ (900 * 7200),

and therefore,

Ek = MOI * RPM2 * 1.523087x10-6 watt-hours (Eqn 4, to six decimal places)

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#3
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Re: Formula to find horsepower hours

02/07/2009 11:11 PM

There's a GA for ya.

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#4

Re: Formula to find horsepower hours

02/08/2009 12:37 AM

This may be off-subject, but I have a request/suggestion. In college I had some subjects that I just "got" and others I had to either memorize or at least deal with on a purely theoretical/mathematical basis. This latter approach took a lot more work, was a lot less fun and ultimately fruitless. My solution was to switch from electrical engineering to mechanical engineering, and for me that worked. BUT - I always wondered if that had to be the case. If I had a heath kit instead of an erector set as a kid would I have developed an intuitive understanding of electronics? Near the end of my first degree I did take an EE class for MEs that took the approach of using analogies, and that really worked well for me.

Anyway, I have a request to those of you who seem to have such a strong grasp of energy/power. Can you frame your answers in the form of a tangible, physical example? For instance, if the robot had a 3 foot arm swinging a 5 pound sledge, what are the forces and energy required to get the sledge up to speed? Also, ssuming one doesn't observe it too closely and get whacked, how much energy will it consume as it swings about? Better yet, how many AA batteries would it take to keep the arm running, taking into account standard efficiency ratings for motors and bearings (personally I have trouble believing air resistance is the major source of loss, but in school we always did assume frictionless bearings and perfect motors and such). This is the whole point of building a robot, in my mind: To make the theoretical tangible.

I hope this makes sense. I am not trying to be cheeky here. I honestly believe there is a difference between learning and truly understanding, and that gap must be battled in every course and every lesson until it no longer exists.

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#5
In reply to #4

Re: Formula to find horsepower hours

02/08/2009 1:46 AM

I do believe you are correct that you learn by doing - if you had played with electronics and gotten "a feel" for it, a EE degree would have worked for you. But the world needs all kinds, so... all's well that ends well.

Regarding your question, a little more info is necessary. What do you mean by "energy required to get the sledge up to speed"? Energy will be proportional to speed squared, so the speed issue is critical. This is physics, by the way; nothing to do with electricity per se. If you are content with a snail's pace, literally, then that AA battery might just do the trick. If you want that mallet swung so that it performs a useful task, sorry, a AA won't cut the mustard.

Here's the physics. In a constant gravitational field, the work you do (energy expended) to lift the mallet a certain height is

E = m*g*h

where,

E is energy (Joules),

m is the mass lifted, kilograms,

g is the acceleration of gravity at the earth's surface (9.8 meters/sec^2), and

h = height the mass is lifted in meters

So if our robot is built like a six foot tall man, with a vertical reach of eight feet, and the three foot sledge handle extends beyond that eight foot reach, then the total height the mallet is raised is eleven feet.

Our inputs to the above equation are:

m = 5 lbs * 0.454 kg/lb = 2.27 kg

g = 9.8 m/s^2

h = 11 feet * 12 inches/foot* 1 meter/39.37 inches = 3.35 meters

So the energy in Joules required to lift the mallet eleven feet is:

E = 2.27 kg * 9.8 m/s^2 * 3.35 meters = 74.6 kg m^2/s^2 = 75 J

Note the dimensional analysis of the calculated energy has units of mass times the square of velocity, just as it should per my original post on this subject.

One final bit of physics before we get to electricity. Note I ignored the mass of the sledge handle. That makes a more complex solution, because the mass of the handle is distributed. You would work the problem as if all the handle mass were located at the midpoint, and calculate lifting the handle mass to the height of the midpoint. More on the handle presently - it gets even more complex.

Now on to what kind of battery will do the job. This involves electricity, but also more physics. Battery energy storage is rated by the amount of current it can deliver per unit time. Typical units are ampere-hours, or for a AA, milliamp-hours. Now amp-hours alone don't tell the story, because amp-hours are not equivalent to energy. But if you multiply amp-hours by the battery potential (Volts), then you're in business.

The basic equation is that electrical power equals Volts times Amps. And power is just the rate at which energy is expended, so you have:

P = V * I, and

E = P * t

Now these equations assume constant values. If they are time-varying, you have to perform integrations, but the basic concept doesn't change.

A AA alkaline battery might have about a 3 A-hr rating. That does not mean it can source 3 Amps for an hour, because this is measured with a lot smaller load, like maybe 100 mA for 30 hours. But the overall energy storage is then

E = 1.5 Volts * 3 A-hr = 4.5 Watt-hours, from the power equals voltage times current equation above.

To get to Joules, we note that from E = P * t, a Joule equals one Watt for one second. So a Watt-hour is 3600 Watt seconds, or 3600 Joules. That means our AA battery stores 3600 * 4.5 Joules, or 16,200 J.

Now way back at the beginning, we calculated that lifting the sledge would only take about 75 J, so it looks like one little AA is good for a lot of swings. But we're ignoring a couple items, one small, one very large.

If we were just going to let the sledge free fall, then the 75 J is all we need getting it up. But most of the time, we try to accelerate the sledge downward well beyond what gravity provides, so we will need extra energy on the way down, as well. That's the small problem.

The big problem is getting enough current out of the battery. Remember that a 3 A-hr rating doesn't mean 3 Amps for an hour. For a AA, a much lower current such as 100 mA is what it is designed for. So we need to figure out what sort of acceleration we need for the sledge, and how much the battery will provide. Physics and electricity, again.

Let's say we want to lift the sledge in one second. That means we need to expend 75 Joules in one second, which means 75 Watts. Now recall that electrical power equals Volts times Amps. Our battery is 1.5 Volts, so from

P = V * I

I = P/V, or

I = 75 Watts/1.5 Volts = 50 Amps (!!!)

Fifty Amps isn't a AA, or even a C or D cell - that's called a lead acid car battery.

Now there is one final complicating factor we've ignored, and that is moment of inertia, as per the original post. Everyone has a gut feel that swinging a five pound weight on the end of a mallet takes more effort than to simply heft a five pound weight in your hand. By the same token, the five pound weight on the end of the handle does a lot more damage when it strikes something than you hitting something at the same speed with the weight in your hand. It's all tied up in the moment of inertia. You have to calculate the power to accelerate the moment of inertia as well as the mass. But we've already decided we aren't going to get by with a AA, so we needn't unnecessarily complicate the calculation further - that's just pure engineering.

And finally, you asked about the energy to just swing the mallet around. If you are talking horizontally, then there is no work against gravity, and once you have expended enough energy to bring the mallet up to speed, after that the only energy required is that to overcome friction and push air out of the way. You are probably correct that swinging a mallet doesn't encounter lots of air resistance, but in the original post the device rotated was some sort of spinning weapon, and I pictured it having a propeller like effect, slicing through air as it were, and therefore doing work against the air, hence my comment that air resistance would be the dominant source of energy dissipation once it was up to speed.

Hope that helps!

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#6
In reply to #4

Re: Formula to find horsepower hours

02/09/2009 8:57 AM

Morepower to ya! I, too, should have switched to mechanical engineering, but never did. I think every school should make an effort to explain electrical phenomena with analogies to the tangible, mechanical world. They did a pretty good job of this at Valparaiso (Indiana) University, which is why I now know that inductors act like masses, capacitors act like springs, and resistors act like friction (this, of course, is the basis of the analog computer, which probably doesn't exist anymore). They even took pains to point out that current running through a wire behaves remarkably like water running through a pipe, if you think of voltage as pressure, current as flow rate, and resistance as pipe friction (which is a combination of the Reynolds number and maybe some other factors which I have forgotten in the last 35 years). The point is that ALL physical systems, whether in the macro-world or the invisible, sub-atomic world, follow the same set of rules.

I can't believe there's anything really intuitive about electronics. We grow up in a world where you can start absorbing physical cause-and-effect from the first moment you open your eyes, or possibly even before birth. Electronics, on the other hand, just sits there, silent and motionless; exotic instruments are required just to observe its behavior. It's a miracle anybody can ever understand it.

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#7
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Re: Formula to find horsepower hours

02/09/2009 9:07 AM

"Electronics, on the other hand, just sits there, silent and motionless ..."

You've never seen a really good short-circuit, ya! Neither silent nor motionless! Check out this substation.

"It's a miracle anybody can ever understand it."

You develop a feel - an intuition about it, so to speak - and you can think of it in terms of analogies, too, like the water pipe. After many years doing electrical engineering, I can look at a fairly complex circuit and in a few minutes get a good idea of what it does and how it works - without instruments. This isn't to say I'm special in some way. Just a bit of experience and intuition mostly.

But don't let me under the hood of your car! It'll never see the road again!

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#8

Re: Formula to find horsepower hours

02/09/2009 10:23 AM

Lets Make this simple shall we.

Take the Killowatt-hours the motor has run and divide by 0.7457.

That will give you the Horsepower hours.

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#9
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Re: Formula to find horsepower hours

02/09/2009 10:40 AM

Your conversion from watt-hours to horsepower-hours is correct. However, once the spinning mass is up to speed, running the motor longer does not contribute to the rotational kinetic energy. The OP is interested in the rotational kinetic energy. The spinning mass will still have this energy when the motor is de-energized. The mass will slow down, of course, as it dissipates its energy through friction (or usage), but once the mass is up to speed, the motor is out of the equation except for its own contribution to the total MOI.

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#10
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Re: Formula to find horsepower hours

02/09/2009 10:54 AM

The posters question is for an hours conversion, not an energy related to hours conversion.

Acceleration is mentioned, but nothing about the, consumption of that energy in relations to functional hours. It appears to me that no matter the speed they are looking for a direct reference to the hours use.

But I am probably wrong in my assumptions.

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#11
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Re: Formula to find horsepower hours

02/09/2009 11:02 AM

The OP asks for the energy (watt-hours or horsepower-hours) used to accelerate a spinning mass to some final speed, up to 2,500 RPM. Watt-hours (and HP-hours) are a measure of energy, not power. Computing the energy directly is equivalent to integrating the input power over the acceleration interval. By calculating the rotational kinetic energy of the spinning mass directly from the MOI and the RPM, the total power input (sans incidental losses) gives this result no matter how long it takes to accelerate the mass to speed. The accleration can even be erratic (so long as it's monotonic), but it doesn't matter. The final kinetic energy at speed represents the total power used to accelerate the mass, neglecting losses.

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#12
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Re: Formula to find horsepower hours

02/09/2009 11:43 AM

You are correct I overlooked the word used. Good thing I dont build space shuttles, just HVAC systems.

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#14
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Re: Formula to find horsepower hours

02/09/2009 5:29 PM

Thank you everybody. All I needed was to know how I may calculate the energy used when accelerating my mass. I understand that once my weapon is up to speed, it should take little energy to keep it there lost by friction and air resistance. So, my SolidWorks program tells me that my weapon should have an MOI of .0605 in the direction that I will be rotating it. Plugging this into our formula in Eq 4 of the second reply yields:

Ek = MOI * RPM2 * 1.523087x10-6

.0605 * 2500 RPM2 *1.523087x10-6

.5759 watt hours

The original purpose of this was to find how much battery capacity I should use by finding out how much energy it would take to spin up the weapon after every hit. We will run our motor at about 22.2 volts, so this should yield 0.5759/22.2=.0259Ah used to spin this weapon up. This seems way too small to me. Am I wrong?

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#15
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Re: Formula to find horsepower hours

02/09/2009 6:07 PM

If you drew 0.0259 amps at 22.2 volts for one hour to spin up the weapon during that time, you would impart roughly 2073 Joules to it at a draw of 0.575 watts from your battery. However, if you revved it up to speed in one second you'd be drawing over 93 amps at 22.2 volts (2073 watts) from your battery during that one second. So, yes, that figure sounds reasonable.

Also don't neglect losses in the motor itself when calculating your energy budget. Motor windings waste power in resistive heating. You already mentioned friction and air resistance. Your motor losses will probably be the greatest of these three.

Also note that most batteries' actual Ah capacity depends on the discharge rate. The higher the current draw, the less the battery's apparent Ah capacity. You might want to run some dynamic tests on your candidate batteries to make sure they don't wimp out on you during times of peak demand - especially toward the end of a battle. Temperature also makes a difference in how well batteries perform. Build margin in your energy budget to accommodate these considerations. We want your bot to kick bot butt, so keep us posted, won't you?

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#16
In reply to #14

Re: Formula to find horsepower hours

02/09/2009 6:39 PM

Just curious: how long do you have to spin the weapon up after a hit? If it is just a few seconds, your battery will have to supply a heavy current during that time, and it will have to supply this current repeatedly. One option you might consider is driving the motor at overvoltage to rev it up fast and then open-circuit the motor when up to speed.

Top-fuel drag racers run pure nitro in engines not really designed for the stuff. They push it all the way to the limit in exchange for rebuilding or replacing the engine frequently. They might get several thousand HP from a hemi, but only for a short time. These babies are just on the brink of blowing themselves apart, and some do.

Think of your motor as expendable; lasting only as long as it takes to win. Drive the hell out of it at higher-than-rated voltage. This also has the advantage that you needn't draw such heavy currents from your battery. They'll get hot, too, so be careful.

Do not overvoltage your motor if it is of the "brushless-DC" type. You can fry the motor's internal electronics if you do that. Use a brush-type DC motor if you aren't already. Gives you much more flexibility.

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#17
In reply to #14

Re: Formula to find horsepower hours

02/09/2009 6:59 PM

One more thing (I keep stepping out to run errands. Sorry). You might also consider using a lower-voltage motor with your 22.2 volt battery if you want to try the "overvoltage" approach. What is your present motor's voltage rating?

You could probably pick up some 6 or 12-volt motors pretty cheap. Stress-test a couple of these by revving them up, delivering your payload, and then doing it again; each time cutting the power to your motor when it's up to speed. Do this until you burn them out. It'll give you a good feel for the motor's limits of performance.

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#18
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Re: Formula to find horsepower hours

02/10/2009 10:49 AM

Actually, I've already ordered a very nice brushless motor. It runs at the 22.2 volts (there is a voltage cap of 24 volts in our competition) that I stated before. The motor states that it can draw as much as 100A continuously. The coolest thing is that this motor only weighs three-fourths of a pound, which is absolutely fantastic in my 15 pound class.

As for the batteries, we are almost certain that we will use lithium ion A123 batteries. This type will have a good capacity (about 2.3Ah, in a perfect world) and can supply huge amounts of amperage (70A continuous and 120A for 10 seconds).

I believe that we now have enough data to find out how long it will take my weapon to accelerate. I haven't done a stress test on the motor itself yet (because I lack the power supply necessary at this point). Therefore, I don't know how much current it may draw on normal conditions. But, just for poops and giggles, could we calculate how long it will take for an MOI of 0.0605 to accelerate to 2500 rpm under 22.2V * 100A = 2220 watts?

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#19
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Re: Formula to find horsepower hours

02/11/2009 3:25 PM

"... could we calculate how long it will take for an MOI of 0.0605 to accelerate to 2500 rpm under 22.2V * 100A = 2220 watts?"

Yes, but you have to make some assumptions. The most basic of these is to assume that your power input remains constant until the motor is up to speed. Under this assumption your motor will go from 0-2500 RPM in 2220/2073 seconds, or about 0.933 seconds.

The power input to the motor will probably not be constant, however, as the motor comes up to speed. How fast it actually does come up to speed in this case has a lot to do with how the motor's internal electronics manage the torque. Internally, BDC motors are really three- or four- or even six-phase motors with the internal electronics juggling the phases.

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#20
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Re: Formula to find horsepower hours

02/18/2009 10:23 PM

I certainly hope you have an opportunity to post footage.

I am following this closely.

I have wagered my son's tuition on your team.

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#21
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Re: Formula to find horsepower hours

03/07/2009 5:53 PM

RoboBots 2009 will be held at the Meadville High School gymnasium on April 4th. This is the third year of Tool City RoboBots. I am co-captain of team FreshMASH (Freshman Meadville Area Senior High School), a team that consists solely of freshman from our local high school. Last year there were 15 schools participating and thirty or so teams, and we expect to see many more this year.

Almost all of the teammates of FreshMASH fought with our bot THRESHER last year as the only all-middle school team; going up against seniors and vocational technical school students. We went 2-2 and had one of the most exciting matches of the day (is there a way to post videos?).

This year, we're using the full-body spinner design in the 15 pound weight class. Super megabyte was a very successful full-body spinner. This consists of one chassis that contains motors, batteries, and all the other guts that go into a combat robot, and a spinning weapon or shell. The shell acts as both the armor of the robot and the weapon. At this point in time, we have almost all of the parts we need and just need to drill and tap holes and put the thing together. We've teamed up with Modern Industries and Kuhn Tool and Die to have some parts CNC machined out of solid blocks of aluminum. I have some pictures below of the parts on SolidWorks and the real thing:

The chassis and shell here have been machined out of solid blocks of aluminum, except for the sides. As you can see, the sides and top of the shell are two separate parts, to conserve weight. The sides, instead of using a solid block of aluminum, have been machined out of a large aluminum pipe. You might be wondering exactly how we plan to drive our weapon. The answer is friction drive. Instead of driving the weapon off of a shaft powered by a motor and gearbox directly, we have a 3.5 hp brushless motor driving a wheel against the shell, much like how a car tire contacts the ground.

Here is the top of the shell, which will mate with the sides of the shell by interlocking tabs.

And, last but not least, are the chassis and sides of the robot. If anyone has a suggestion of how to post videos, I would really like to post some footage.

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#22
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Re: Formula to find horsepower hours

04/06/2009 5:37 PM

The competition was last Saturday, April 4th. My team went 4-2 and tied for fourth place out of 28 teams competing. We also won the awards for the Best Engineered Robot, Coolest Robot, and Best Sportsmanship (for hosting the event). I should be getting some videos of our fights soon. Is there any way to pose videos?

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Re: Formula to find horsepower hours

04/07/2009 4:51 PM

Are you the Sam Tome that is looking to further your education? If so, you apparently entered a bogus email address and phone number (mine) on the internet that caused me a stream of wrong numbers today. I got calls from universities trying to contact a Sam Tome. I thought the phone number might have just been a mistake until I asked one of them to email you and let you know the phone number was incorrect and she said the email address was bogus as well. If you wish to do that again, I do hope you will chose to use all 9's or something that will not inconvenience an innocent person.

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Re: Formula to find horsepower hours

04/07/2009 6:03 PM

I am very sorry for this inconvenience. I have never entered such information or even sought to further my education over the internet, although I do get several spam messages every day for such things.

Once again, I am very sorry; we all know how annoying telemarkerters and wrong-numbers can be. But I can assure you that I have had nothing to do with this.

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Re: Formula to find horsepower hours

04/08/2009 6:23 PM

Thank you for replying. Someone may have used a bogus name as well. The calls seem to have stopped. Have a great day!

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#25
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Re: Formula to find horsepower hours

04/07/2009 6:07 PM

You can always upload them on youtube and post a link. I'm anxiously awaiting to see these battles!

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#27
In reply to #22

Re: Formula to find horsepower hours

04/08/2009 8:02 PM

Our local newspaper, The Meadville Tribune, created a slideshow of various pictures they took at the event. Here is the link. Slides 13 through 24 all feature our team and robot, and my hands are in slide 26.

The team captain's sister took videos of every one of our fights; once they upload them onto youtube I'll paste the links here. It was quite destructive.

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Sam Tome
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Join Date: Feb 2009
Posts: 8
#28
In reply to #27

Re: Formula to find horsepower hours

02/14/2010 1:57 PM

Here are the links to the videos on youtube. The link is to our team's channel.

http://www.youtube.com/user/MASHrobotics

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Sam Tome
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#13

Re: Formula to find horsepower hours

02/09/2009 2:40 PM

What will the force be when the other bot receives a whack from the sledge hammer?

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