Flist! Help!
Apr. 30th, 2009 11:47 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
ljgeoffI've got this hazy idea that I need to talk over with Mike -- start with go cart plans, and add pneumatic engines from used air powered drills to provide power to an axels.
But I have no idea how to figure out horse power, rpm, torque, necessary psi or the size of the air canister.
The drill specs list it's "air consumption" at load as 4, but I have no idea of what this means, other than it uses 4-somethings of air per something. Ah, here it is; it's cubic feet per minute. Oh, that chart is handy: each drill would require a tank at 90 psi and would use about 8 cubic feet per minute.
So, an average scuba tank holds 80 cubic feet of 3000 psi air. But what does that mean?
edit: Also looking at these pneumatic motors -- and talked briefly to Mike. I'm trying to understand the payoff between rpm and torque, that is, between how fast it goes and how much it can carry.
If the diameter of a tire is 9", then 60rpm*9inches*3.14*60min/1hr*1mile/63360inches= 1.81 mph
that's with the maximum 903 inch pounds (whatever *that* means)
With the lower torque (345 inch pounds,) I get 500rpm, which would give me 13.38 mph.
This drill goes to 2600 rpm, but it rates as .75 horse power, instead of a torque rating. So if I had a motor on each wheel, would that be 6 horse power, with a max speed of 70 mph?
Ok, here's a explanation of torque and horse power, for anyone interested:
For purposes of this discussion, we need to measure units of force from rotating objects such as crankshafts, so we'll use terms which define a *twisting* force, such as foot pounds of torque. A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum.
Now, it's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.
Visualize that one pound weight we mentioned, one foot from the fulcrum on its weightless bar. If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidently, we have done 6.2832 foot pounds of work.
OK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement:
Torque * RPM
Horsepower = -----------
5252
The specs for my Festiva says that it has 63 horsepower at 5000rpm. It has a curb weight of about 1800 lbs, and can haul around an additional 1000 in passengers and stuff. Does that mean one Festiva horsepower can carry 44 pounds? Does that mean that my 3 hp go cart will be able to push around just 132 lbs? Well, that won't work. Hmm.
But I have no idea how to figure out horse power, rpm, torque, necessary psi or the size of the air canister.
The drill specs list it's "air consumption" at load as 4, but I have no idea of what this means, other than it uses 4-somethings of air per something. Ah, here it is; it's cubic feet per minute. Oh, that chart is handy: each drill would require a tank at 90 psi and would use about 8 cubic feet per minute.
So, an average scuba tank holds 80 cubic feet of 3000 psi air. But what does that mean?
edit: Also looking at these pneumatic motors -- and talked briefly to Mike. I'm trying to understand the payoff between rpm and torque, that is, between how fast it goes and how much it can carry.
If the diameter of a tire is 9", then 60rpm*9inches*3.14*60min/1hr*1mile/63360inches= 1.81 mph
that's with the maximum 903 inch pounds (whatever *that* means)
With the lower torque (345 inch pounds,) I get 500rpm, which would give me 13.38 mph.
This drill goes to 2600 rpm, but it rates as .75 horse power, instead of a torque rating. So if I had a motor on each wheel, would that be 6 horse power, with a max speed of 70 mph?
Ok, here's a explanation of torque and horse power, for anyone interested:
For purposes of this discussion, we need to measure units of force from rotating objects such as crankshafts, so we'll use terms which define a *twisting* force, such as foot pounds of torque. A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum.
Now, it's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.
Visualize that one pound weight we mentioned, one foot from the fulcrum on its weightless bar. If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidently, we have done 6.2832 foot pounds of work.
OK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement:
Torque * RPM
Horsepower = -----------
5252
The specs for my Festiva says that it has 63 horsepower at 5000rpm. It has a curb weight of about 1800 lbs, and can haul around an additional 1000 in passengers and stuff. Does that mean one Festiva horsepower can carry 44 pounds? Does that mean that my 3 hp go cart will be able to push around just 132 lbs? Well, that won't work. Hmm.
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(no subject)
Date: 2009-04-30 08:39 pm (UTC)See if you can find an application needing an engine (air conditioner? water pump? conveyer belt?)... that can be solidly mounted to its base while doing its work. When you put an engine on a vehicle, you vastly expand the number of technical glitches you can expect to encounter.
This way, if the engine blows up sending shrapnel everywhere, at least it's in a specific, containable space where you can control access.
Once you've perfected your mad science invention, you can install the mark 2 version onto a snowmobile, ot hovercraft, or....
(no subject)
Date: 2009-04-30 09:06 pm (UTC)Oh, and if you want to get all renewable energy about it, us a windmill to compress the air.
(no subject)
Date: 2009-04-30 09:31 pm (UTC)Yep, that's the plan (http://the-stronghold.livejournal.com/3521.html).
(no subject)
Date: 2009-04-30 09:42 pm (UTC)"I also made an air powered blender, I took the motor from a cheap harbor freight die grinder and fit it in a cheap electric blender with a quick connect on the side - it fit perfectly, it just took a couple of hose clamps to hold it. Plug it in to the mog and bingo - easy margaritas... The problem is the cheap die grinder motor takes a lot of air at 100+ psi, it'll empty my tanks really quick if the engine is not running and if you don't run an air motor hard they have very little power. I think the motor I have uses something like 7 or 9 cubic feet a minute but more expensive grinders use a lot less. Another thing to watch out for is the motor has so much torque it will crack the plastic casing." Thread is here (http://www.expeditionportal.com/forum/archive/index.php/t-3319.html).
(no subject)
Date: 2009-04-30 09:50 pm (UTC)(no subject)
Date: 2009-04-30 10:25 pm (UTC)"Because PULSAIR uses compressed air or gas to operate, it is helpful to understand how the volume of compressed air is calculated.
A standard cubic foot of air is the amount of air in one cubic foot of space at 70 degrees F and 14.7 PSIA, i.e. standard atmospheric conditions. Since pressure gauges do not register atmospheric pressure, 14.7 PSIA equates to 0 PSIG. Because this is approximately the way air exists in nature, air at atmospheric conditions is also called "free air". Therefore, 1 Standard Cubic Foot of air is the same as I cubic foot of free air.
A tank with a volume of 4 cubic feet holds 4 Standard Cubic Feet of air at atmospheric pressure (0 PSIG). If we inject air into the tank until the internal pressure is 44.1 PSIG, we have 4 cubic feet of air at 44.1 PSIG. How many Standard Cubic Feet of air is this?
Answer: Divide 44.1 by 14.7 = 3. This means that at 44.1 PSIG each cubic foot will contain three atmospheres plus the original 1 atmosphere that existed before pressurizing, or a total of 4 atmospheres. Multiply 4 cubic feet of volume by 4 atmospheres = 16 Standard Cubic Feet of air.
If we inject more air and double the pressure to 88.2 PSIG, the same math tells us we have 28 Standard Cubic Feet of air in the 4 cubic foot tank (88.2/14.7 = 6 atmospheres + 1 atmosphere = 7 atmospheres x 4 cu. ft = 28 SCF).
By using this standard unit of measurement we greatly simplify discussions and calculations of air volumes."
Additionally (http://www.ppi2pass.com/forums/posts/list/1155084317.page), "psia is the absolute pressure. psig is the gage pressure. gage pressure is measured in reference to the atmospheric pressure. For example if the atmosphere has a pressuer of 10psi and the tank had an internal pressure of 15 psi higher than the atmosphere, the psia = 25 psi and the psig = 15psi."
So, if I use a 50 gallon hot water tank as my air tank, I'll have about 6.7 cubic feet (1 US gallon = 0.133680556 cubic feet), and if I pump air into it until the psig is 90, I'd divide 90 by 14.7= 6.12
So, that's 6.12 atmospheres times 6.7 cubic feet = about 41 cubic feet plus the original 6.7 cubic feet comes to about 47.7 cubic feet.
I wonder how much psi a water tank can hold?