Static versus Dynamic Ergs – what’s the difference?


There are a lot of discussions online about dynamic and static ergs (indoor rowing machines).  We clarify using physics and some basic engineering what the differences are.

The fundamental difference between the mechanics of a static ergometer (such as Concept 2) and a dynamic ergometer (or a boat on water) can be found by taking the following test.

  • Place yourself at front-stops and exert pressure on the footstretcher, you move backwards in proportion to the force exerted.
  • However if you sit at the frontstops in a boat and do the same, you only move backwards relative to the bank by an amount ~20% of your leg length. The rest of the motion is taken by the boat moving away from you.

This is a result of Newton’s 3rd law of action-reaction. The force exerted by your legs on the flywheel acts equally on both you and the flywheel. In the case of a static erg, the flywheel remains stationary as it is mounted to the ground and instead you do all the moving. In case of a dynamic erg (or a boat on water), the mass of the boat is typically 10-20% lighter than you, making it mover further than you.

In the static ergometer you are actually putting in more effort to accelerating your body weight compared to the dynamic ergometer, which splits the acceleration of your body weight and the boat in opposite directions.

A dynamic erg such as Rowperfect, attempts to simulate the reaction effect by having the flywheel moving on the rail to absorb most of the motion. The Rowperfect weighs approximately the same as a sculling boat providing you with an experience of rowing and not sliding.

The designer of Rowperfect Casper Rekers performed a test comparing the power output for an erg with and without a fixed flywheel. The test results showed the subject gaining about 10-20% power output without a fixed flywheel. This additional power could be used to accelerate the flywheel instead of one’s bodyweight. This is also one reason why the catch of the static erg feels ‘slack’ compared to the boat.

To refer to the maths behind these principles click here.

Other articles about static versus dynamic ergs

  1. Static versus dynamic ergs – Concept 2 response
  2. Five reasons why I choose a dynamic erg
  3. The Catch on dynamic erg versus static erg
  4. Discussion on Rowing Illustrated Dynamic Erg – Jan 2011

Enhanced by Zemanta


Subscribe to our Newsletter for more information:

Our Rowing Network

Did you like this post? Support this blog and our network by donating

Feel like Rowing again? Visit the Shop at!

Accessories, equipment, boats, clothing

This Post Has 4 Comments

  1. Andy Millar

    The “test” and maths/physics arguments are interesting enough, but of course, artifical as they are based on a system with oars out of the water -thus removing the fulcrum on which the propulsive force relies. It seems to be a generally safe to assume that a lever without a fulcrum does not behave (nor can be analysed) in the same way as with a fulcrum in place………….albeit with a proportionately small amount of movement in the fulcrum in the case of rowing

  2. Colenso

    “This is a result of Newton’s 3rd law of action-reaction. The force exerted by your legs on the flywheel acts equally on both you and the flywheel.” ~ Mohit Gianchandani

    This is not correct. Forget the so-called “action-reaction” formulation of Newton’s third law. Everybody who tries to express Newton’s third law in terms of action and reaction invariably fails to understand or apply the law correctly. (It seems to be an engineer’s mantra and all mathematical physicists know that engineers do not really understand the principles of Newtonian mechanics – they just hope they do!)

    Instead, it is much more helpful and precise in all situations to start with the formulation that Newton’s third law of motion states that all forces exist in pairs. This means that if a system A acts with a force F(ab) on a system B, then system B acts with an equal and opposite force F(ba) on system A. In other words, F(ab) = -F(ba), and | F(ab) | = |-F(ba)|. That is, the magnitude of F(ab) is identical to the magnitude of F(ba), but the direction is opposite. (All forces are physical vector quantities. To specify a physical vector quantity, we must specify its magnitude and its direction.)

    Note also that fully labelled diagrams are essential in all discussions of classical mechanics. If you cannot draw it then you do not understand it. Fully labelled diagrams quickly reveal mistakes in understanding. Unfortunately, I cannot provide a diagram here so must rely on mere words.

    So, back to the example. In the case of my feet together pushing on the erg’s foot-piece with a force F(fp), the erg’s foot-piece pushes back on my feet with the force F(pf) where F(pf) = – F(fp). (NB f designates feet; p the erg’s foot piece.)

    Lastly, although it may be pedagogically convenient for school teachers and introductory text books to start a discussion of classical mechanics by talking about forces, together with motion in a straight line, in the real world we are concerned almost always with torques, together with motion which is not in a straight line – ie it is non-rectilinear.

    Therefore, we need to replace Newton’s classical statements for rectilinear motion with the broader statements of principle that cover the case of all motion, non-rectilinear as well as rectilinear, in classical systems. Generalised in this way, Newton’s third law then may be restated as: All torques exist in pairs. This means that if a system A acts with a torque T(ab) on a system B, then system B acts with an equal and opposite force T(ba) on system A. That is, T(ab) = -T(ba), and | T(ab) | = |-T(ba)|(The magnitude of T(ab) is identical to the magnitude of T(ba), but the direction is opposite). (One may also see from this that Newton’s third law for forces is simply a special case of Newton’s third law applied to torques.)

  3. Colenso

    Correction, should have written:

    “This means that if a system A acts with a torque T(ab) on a system B, then system B acts with an equal and opposite torque T(ba) on system A. “

    1. Rebecca Caroe

      Colenso – sounds like you are more expert than I am! Thanks for the comment.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Related Posts

Whether it’s sports, nutrition or equipment, you’ll find what you’re looking for in one of these categories


Everything about rowing – browse through our large archive or search for articles of your choice


Search Blog


Rowing Network

Do you like our posts? Support this blog and our network by donating


Get all latest content and news!