Jump to content
Mechanical Engineering

G B Reid MIMechE, SIMarEST

  • Posts

  • Joined

  • Last visited

  • Days Won


G B Reid MIMechE, SIMarEST last won the day on December 19 2019

G B Reid MIMechE, SIMarEST had the most liked content!


Profile Information

  • Gender
  • Location
    Benfleet, England
  • Interests
    [NOTE: No longer SIMarEST, but connot remove!] Mechanical engineering "generalist", experience in Design (industrial, mechanical, aerospace grade components), Rail, Underground, defence, process flow optimisation, Project Engineering and Mechanical Systems 3D CAD user (various) since 1992. Electrical design installation and verification.

Recent Profile Visitors

3,974 profile views
  1. They work just like the 1980s cartoon series "Battle of the Planets"....G-Force!
  2. I shan't give you the answer, as that is doing the job for you, but I will pont you in the right direction...(besides, there is sufficient ambiguity in your description to make this rather awkward without lots of clarification) If you ignore compressability, model this as a "Constant Volume", you know the diameters of the pipes, and you also know how much total area the outlets are...the back-calculation is very simple from there! Good luck
  3. This made me Smile! As I said in my post, I have discussed this and had the detail in the "Seapower" sketches expanded on in the past, so I understand where the diagram is coming from and the idea itself...it is a little "unclear" from the above but the concept IS the "Seapower" diagram, the umbrella, a potential drive facilitator for the bubbles.
  4. I've just perused this again and had a thought....you're making a punch...have you calculated whether the force required to punch through the metal can be achieved? It's a CSA calculation (circumference x thickness) and the shear strength of the material in Pa Take the CSA (in m^2) and multiply by the shear strength in Pa to get the required force in N. Good Luck!
  5. Hi Govardhanan! You will likely have problems with this design due to the foodstuffs not flowing sufficiently. If the foodstuff is too viscous, it will not exit through the holes at the bottom...in fact in some instances, the container will fail before it does. If you look at dispensers available commercially, you will find that the majority have a cone at the bottom to facilitate exiting of the material. Whilst you acknowledge that you are a student of mechanics, I am sure you will be familiar with the phrase "every action has an equal and opposite reaction." (Isaac Newton) . Basically, in this situation, all of the force from the piston is reflected back into the piston with nothing facilitating flow through the exit holes. If this exit was altered to a cone of - say - 60 degrees (radially), 86.6% of the force would push down and 50% of the force would be deflected to "feeding" the material through towards the target exit. [Yes that does exceed 100%, but that's trigonometric mechanical advantage] Note: I have not sketched this out and checked as have no paper at the moment, so it's a "finger in the air" calc. but at very least the principle is right if the figures are wrong... Good luck! Bruce
  6. Failing the accelerometer - which would be very successful if on a drum - you could connect a microphone to a transistor to switch on the light. It is a very simple circuit and is reproduced on a number of differing electronics websites. Basically the "switch" would be the microphone. They generally have a resistor in the circuit...if this was a variable resistor, it could be adjustable. daisy chain this to a timer circuit and you're laughing....or drumming! Good luck!
  7. I think the best option is to ask "what sort of buoyancy force could this generate....?" The buoyancy gives the force the speed of which would be limited hydrodynamically - basically by the speed of the bubbles rising. Asking "Torque" is not really helpful as this will depend entirely on the size of the cog used to drive the unit. Basically Torque is Force x distance. You need to first know the force and the systems can be modified for the torque sought....the speed you have already calculated. You have done most of the work on this....the buoyancy is identical to "The sum of" "the volume of the bubbles x the mass of water x g"; you have already calculated these at various depths: Basically, calculate the volumes, add them up and multiply by 1000kgm^-3 and then 9.81ms^-2 (Metric). This would calculate down and cancel to give "kg m s^-2" - which is Newtons - the fundamental unit of force in SI units. I apologise, we have communed on this in the past, I just haven't had the time to look into this in detail, and time I would need. I confess to being a "Metric only" guy, which basically means I'd have to spend the time recalculating everything to make head nor tail of it and verify your original calcs (which I would not convert, but redo to check). [Have to confess - I took your name, "tooldtocare" as being "tool'd to care"....I guess I must be a cup half full guy!🤣] Hope this helps Bruce
  8. This answer is from a kinematics perspective - if you are querying something other, please be clear in your query. In a system, if there is 1 DoF; there is 1 DoF. The "wetness" of the surface affects the coefficient of friction (mu) and is a non-dimensional multiplier...it will therefore have no effect on the number of degrees of freedom. I hope this helps.
  9. <sigh>...I'll give you a clue...you're looking for "dI/dt"... (note: that's not an "L")
  10. The most likely reason is "fatigue" (high quality close-up images of the broken surface could confirm this). If you bend a bit of metal back and fourth, it will snap...that's an extreme example of fatigue. Fatigue happens when an item is loaded and unloaded multiple times and can cause items to fail when there is very little actual use due to the cyclical nature...it is effectively crack propagation within the lattice structure of the metal's crystalline structure. [its not even "real" cracks, it is incomplete crystals shunting individual atoms across internally within the lattice...they multiply until they become significant.] It is a little complex in reality, but a general rule of thumb is if the item subject to fatigue is designed to Load/0.3, it should significantly increase the lifespan....and use the yield point for the calcs, not the UTS. Good luck Bruce
  11. Hi Pablo Alas, I have no first hand experience with the Jominy, so I will not comment on this. (I only comment where I have knowledge or first hand experience...) That said, on the other, yes, use the figures for the 100mm, if incorrect, the error will be on the "safe" side. Bruce
  12. Hi AliceIn....apologies for the delay, just read your message! ALWAYS err on the side of caution and use the lowest value unless absolutely sure! Warmest wishes Bruce
  13. That's a loaded question...(please excuse the pun!)...it depends! If you are building a structure subject to repeated loading, or safety critical, you should use the yield...NOT the UTS. If you are designing fastners, the UTS is more relvant...but then again, if subjected to vibration or "impulse" loading, again the Yield becomes far more important. It is not a straight forward answer to what seems a simple question. The differring UTS values are generally due to the heat-sink aspect of the material undergoing heat treatment. Basically when an item is hardened (quenching for steels). The item is heated up passed the crystalisation temperaure and then quenched in either oil or water. This takes the heat out of the material so quickly that the crystal-growth is restricted and they remain small. The thicker the item, the more heat remains within the material allowing the crystals to grow to larger (unwanted) sizes....basically for hardness, smaller crystals are better (in steels al least - be careful...other metlaic materials differ quite significantly!) I would always err on the side of caution and use the lowest value. This is especially the case for a shaft...the hardening may only affect the surface! For the diagram....for me there is insufficient detail to answer the question....there would be accompanying information to expain further what it is showing...the only answer I can give is that it is two differing samples! (although...as I think about it further, it is possible my previous answer may actually be helpful here!)
  14. Just to expand a little on Dudley's answer... Professional thread dies are adjustable and used alongside wire gauges to determine the true size of the cut thread. These are NOT the same as the units readily available for hobby engineers. Basically the wire gauges engage with the thread INSIDE the cut thread and are measured across the cut thread....measuring the effective thread size. They can be adjusted to ensure that the thread is correct at the final size. Hope this helps! Bruce
  • Create New...