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Exercise equipment w/ built-in scale

Have machine measure user's weight without requiring manual entry
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Many exercise machines require the user to enter their weight in order to determine caloric usage. Since the number of calories burned on e.g. a stairclimber will be roughly proportional to the user's weight, it's necessary for the user's weight to be entered accurately for the calorie count to be correct.

It seems a waste to require the user to enter such data, though, when for minimal added cost the machine could measure it. Indeed, for some machines like stair climbers, the electronics in the machine should be able to determine the user's weight without requiring any extra sensors (a typical gym-style stair machine uses a generator, resistor, and switching logic as a dynamic brake to set the level of workout; knowledge of the speed, motor, and resistor characteristics would allow computation of the current and power produced. From these, weight could be determined easily.

Wonder why I've not seen exercise equipment do that?

supercat, Mar 07 2002

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       I was viewing the scale not as a precision device which informs the user how much weight was gained or lost during, or as a result of, a particular workout, but rather as a device which could measure the person's weight accurately enough for proper calorie computation. Your point about drinking water to recover from exercise and "gaining" weight is a cute one, though.
supercat, Mar 07 2002
  

       Hey, how about if the scale mechanism could be set to report the delta between when you started and when you stopped excercising? Maybe just the delta so you wouldn't get all depressed about your absolute weight.
bristolz, Mar 07 2002
  

       I could see measuring before or after using the equipment, but not during.
phoenix, Mar 07 2002
  

       Phoenix: On a stair machine, the user generates power proportional to (weight times speed). Since the machine knows the speed, and since the machine knows how much dynamic braking must be applied to maintain the set speed, it should therefore be able to compute power. From power, it could thus if desired compute weight. Though for calorie computations power is what really counts.<p>   

       As a simple example, suppose that the stair machine has a generator whose gearing and windings are such that the generator yields 100 volts per meter-per-second, and 10 milliamps per newton. In series with the generator, the machine has a resistor and solid-state switch. The combined resistance of the generator windings, switch (when 'closed'), and external resistor 1 ohm.   

       If the switch were closed all the time, a 1,000 newton (225lb) individual would fall at a rate of 0.1 meter/second. If the switch were closed half the time, the individual would fall twice as fast (0.2m/s); if a third of the time, three times as fast (0.3m/s), etc.   

       If the individual only weighed 500 newton (112lb), the for any given duty cycle of switch closure, the individual would fall half as fast as a 1,000 newton individual. Likewise if the individual weighed 1,500 newton (337lb) the platform would fall half again as fast as with the 1,000 newton individual.   

       Suppose the machine is set for a descent rate of 0.5m/s (about 1mph, or about 6 seconds per "floor"). It would vary the duty cycle of the switch to try to maintain that speed. With a 1,000 newton individual, the switch would be closed 1/5 of the time. With a 1,500 newton individual, about 3/10. With a 500 newton individual, about 1/10. Since the controller has to vary the duty cycle to regulate speed, it should know how much the person weighs and how much energy is being produced/dissipated.
supercat, Mar 08 2002
  

       (Wonders if [supercat] and [Vernon] are related)   

       You're assuming the body is in freefall (not effected by external stimulus), which it won't be. The person's body will be shifting some combination of right, left, forward and reverse. The person may be pushing or pulling. Part of the person's body may be at still while the rest is in motion. The person may be holding on to handrails. All these things will effect the calculation.
phoenix, Mar 08 2002
  

       A newton is, as you note, a unit of force. Any object in the immediate vicinity of Earth will be pulled toward it with a force of 9.80 newtons/kilogram. While people are more commonly weighed in kilograms rather than newtons, I used the latter measurement for two reasons:   

       -1- Energy is measured in newton-meters (joules), and power in newton-meters-per-second (watts). Converting the person's weight to newtons at the start avoids numerically-annoying conversions later.   

       -2- If the stair machine were used in different gravitational environments (say someone installed on in a moon base), the amount of work the person would be doing would be proportional to the user's weight, not their mass. The 102kg individual mentioned before (1000N on Earth) would only have to do 1/5 as much work on the moon (where he'd weigh about 200N). The machine would register him as weighing 1/5 as much even though his mass was exactly the same as on earth.
supercat, Mar 08 2002
  

       I'm impressed that so far no one's been tempted to take account of the weight (or mass) loss caused by expenditure of energy implied by E=mc².
hippo, Mar 08 2002
  

       phoenix: When a stair machine is properly used, the user will constantly be attempting to raise himself while the steps/pedals are constantly falling. Use of handrails during exercise is not allowed. While the person's center of mass may move slightly during exercise, the dominant effect is that of the person pushing against the pedals with a force equal to his weight, trying to push himself up as quickly as the pedals are falling.   

       I am assuming, btw, that the stair machine is one of the designs with two pedals which are only supported by a weak return spring and a 'ratcheted' dynamic brake/flywheel, where the pedals move down at a constant speed when the user is standing on them. Such designs have the advantage that the level of exercise is nearly independent of the user's step rate provided the user doesn't let the pedals 'bottom out'. If the user takes more time between steps, larger steps will be required than if the user steps more frequently.
supercat, Mar 08 2002
  

       I was right, it's [Vernon].
phoenix, Mar 08 2002
  

       Side note (or at least it is at this point in the discussion :) Calorie consumption is not based solely on the exerciser's weight.
jester, May 02 2002
  
      
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