On treadmill exercise machines, what are watts and mets?
451714
Watts is a unit of energy. You probably have 100 watt light bulbs in your house. You probably also have 60 and 75 watt bulbs as 100 is pretty bright. If you are running and creating 100 watts you could power a 100 watt bulb--most peole can't do this for long. I don't remember what METS are, but it is some sort of power output also.
621094
20 minutes of treadmill walking at 4mph and 15% incline (my favorite!!!) burns 249 calories\n20 minutes of treadmill walking at 4mph and 5% incline burns 137 calories
1104997
run... actually... treadmill workout: do incline walk for an hour... make sure the incline is all the way to the highest (for the gym i go to 15 is the highest incline) and use 3.5 - 4.0 speed.   then stairmaster and randomize it.  don't do bikes cuz it doesn't do anything.  SO every other day, do stairmaster and the other days, incline walk.  make sure you get one day off though.  good luck!
909251
Treadmill always good to change things up:\n\n5 min.\n 3.0 mph - warm-up\n\n\n5 min.\n 3.5 - 4.5 - walk/slow jog\n\n\n1 min\n 5.0 - 5.5 - speed walk/run\n\n\n2 min\n 4.0 - 5.0 - walk/slow jog\n\n\n1 min\n 5.0 - 5.5 - speed walk/ run\n\n\n2 min 4.0 - 5.0 - walk/slow jog\n\n1 min\n 5.3 - 5.7 - speed walk/run\n\n \n2 min\n 4.0 - 5.0 - walk/ slow jog\n\n\n1 min Walk or run as fast as you can\n\n2 min 4.0 - 5.0 - walk/ slow jog\n\n5 min 3.0 - 4.0\n\nI'm not sure about the weight bench maybe some different workouts for abs, arms, legs such as:\n\nplacing your hands on the edge of the weight bench lower your body so it almost touches the floor and than lift up again. \n\nor lay on your stomach with your feet on the weight bench and do some push ups.\n\nSimple stuff like that with the weight bench might work.
277835
CMEs are measured in watts per metres squared.
237451
Hi! I have one of these. It is a little weird at first. I thought I was going to kill myself on it! However, with very little time you can get the hang of it.\nIt is a pretty good workout but isn't that intense. \nI prefer a treadmill.
433894
This is a good question because I also use this machine at the gym. From what I have read, the eliptical macine is only effective at counting calories when used on the higher resistance settings. That may be the reason you find it so easy to use. I usually set it to 14 and do that for 400 calories. Then move on to the tread mill for another 400 and/or the stepper machine. I find that in order to get a good workout, multiple machines should be used othrwise your body will get used to that one machine and not be as effective for targeting all muscles in your body. Heres a litle fun fact...in order to "burn" a pound of fat, you have to burn a little over 3000 calories. Have fun at the gym and you will have a great workout.
452470
That entirely depends on your goals.  If your goals is to increase cardiovascular endurance, then the best machines for you are probably ellipticals, treadmills and stationary cycles.  If your goals are to increase strength and muscularity, then free weights, hammer strength and different machines that rely on resistance are more advisable.  \n\nI personally do not think there is one "best exercise machine" but rather a diverse set of tools is best to accomplish a specific goal whether cardio, stength or power.
1046913
treadmill most definately....works more of the legs and butt than a bike does
1015451
Power = work done per unit of time.\n\nP=W/T
63294
Watts = Volts * Amps\n\nSo if you have amps, multiply it by the voltage (if it's a standard electrical device for the home, the voltage is 120 Volts) to get watts.\n\n1000 Watts = 1 kilowatt
142941
What kind of treadmill you need depends on several things:  who is using the treadmill, how much you are using the treadmill, what you are using the treadmill for (walking or running), and weight of people using the treadmill.  A good website for evaluating these factors and treadmill reviews is http://www.treadmilldoctor.com  They don't sell treadmills themselves, so they are objective.
903919
Generator sizing becomes more difficult when electric motors are involved, due to their large starting demand.  Typically, there are two motor styles: CODE G requires 3 times running watts to start, and CODE L requires 5 to 6 times running watts to start.\n\nIt is not practical to think that all loads will start at the same time.  Nor is it practical to size a generator using all starting watts for all selected motors, as some motor loads will already be running (or OFF), when others are ready to start.  As you can see, it can be difficult to exactly size a generator for a group of specific loads, as they are constantly changing in their "ON" or "OFF" cycle.  Always try to limit your load selections to the bare essential items.\n\nTO FIND:\n1 PHASE KW = V * A * PF / 1000\n3 PHASE KW = V * A * 1.73 * 0.8 / 1000\n1 PHASE KVA = V * A * 1000\n3 PHASE KVA = V * A * 1.73 / 1000\nMOTOR - REQD 1 PHASE KW = ( MOTOR HP) * (EFF)\nMOTOR - REQD 3 PHASE KW = (HP) * (0.746) * (EFF)\nMOTOR - REQD 1 PHASE KVA = (MOTOR HP) * (EFF) * (0.8)\nMOTOR - REQD 3 PHASE KVA = (HP) * (0.746) * (EFF) * (0.8)\n1 PHASE PF = 3 PHASE PF = KW / KVA\n1 PHASE AMPS = (KW * 1000) / (VOLTS * PF) =  KVA * 1000 / VOLTS\n3 PHASE AMPS = (KW * 1000) / (1.73 * VOLTS * PF) = (KVA * 1000) / (1.73 * VOLTS)\n \nREQUIRED PRIME MOVER (ENGINE) HORSE POWER\n KW / (GENERATOR EFF. X .746)\n \nFREQUENCY (HERTZ)\n NUMBER OF POLES X R.P.M / 120\n \n ~ ~\nEXAMPLES IN GENERATOR SIZING FOR LARGE ELECTRIC MOTORS \n\nAll AC electric motors require large amounts of electric power to start up, due to it's starting winding. After approximately 3/4 - 1-1/2 seconds, the starting winding drops out, and the running winding continues, at a much lower power demand.  A generator must be sized to handle the load of the starting winding.\n\nSizing a generator for three phase motor starting LRA : \nThe formula is: Motor HP x KVA per HP x (1000 / (Volts x 1.73)) = LRA. \nGiven the size of your machines, estimate KVA/HP at approximately 6.\nTherefore, for 11 kW ... LRA = 134.6 = 135 amps. Generator must produce this current at 380 V, so (1.73 * V * A * 0.8 / 1000) = KW = 71.\nSimilarly, for 15 kW .... LRA = 184 A and KW = 97.\n\nBOTH SINGLE PHASE AND THREE PHASE MOTOR SIZING EXAMPLES ARE BASED ON INFORMATION IN NATIONAL ELECTRICAL CODE HANDBOOK, ARTICLE 430.7 AND IS INTENDED FOR MAXIMUM VOLTAGE DIP OF 25% UPON INITIAL MOTOR START-UP.  IF 35% VOLTAGE DIP IS ALLOWABLE, REDUCE LRA, THEREFORE KW SIZE, BY 25% (LRA or KW X .75). IT IS NOT RECOMMENDED TO CHOOSE ANY GENERATOR KW SIZING THAT YIELDS MORE THAN 35% VOLTAGE DIP.\n\nStarting more than one motor, at one time: make a list of all motors, totaling all running watts.  Calculate the starting demand when the largest motor is started while all others are running.  Add the largest motor starting watts, to the running watts of all smaller motors already running.
346519
This question is a little tricky. Work is the product of FORCE X DISTANCE. Work is energy that has been used or transferred. Work is accomplished when force acts through distance. For example, when a force of 1 pound acts through a distance of 1 foot, 1 foot pound of work is accomplished. \n\nI will have to go with Answer D,  it would be a better chose if the question was about something like motor efficiency. The efficiency of any type of machine is the ratio of the output power to the input power. This ratio is commonly expressed as a percent. Because all machines have some losses, the efficiency will never be 100 percent. \n\nEfficiency = (output/input) X 100\n\nEfficiency = [output / (output + losses)] X 100\n\nor\n\nEfficiency = [(input - losses)/ input] X 100
479446
treadmill would lkely burn 30% more calories for average users, but this would depend on intensity.\nIf boht users achieved similar target heart rates, and maintained them , the calorie consumption would be similar.
189282
I really like to use the row machine or the escalator like stairmaster for 15 minutes.  They get my heart rate up and my muscles warmed up nice.  Then, I usually run or the use the elliptical for 30 minutes and finish off with a 5 minute cool-down.
762388
take  g = 9.8 m/s^2  and let us assume the load is being lifted vertically upwards. This should have been mentioned in the question or the force against which it is being moved should have been mentioned. In the absence of that we make these simplifying assumptions.\n\nThe motor does 45x9.8x200 joules of work in 90 seconds so the power with which the machine works is \n\n(45x9.8x200)/90 = 980 J/s or 980 watt