Calculating RPM drop with different size propellers

[Viper1953]

Hi all,

So, I've been researching alot more, and found a lovely little article here: http://thequadcopterguy.blogspot.co.uk/p/choosing-your-parts_23.html

It seems extremely useful for first timers like myself who wants to do the calculations as accurately as possible.

However, there is one statement which has me confused:

"As we see above our motor runs at 6660 RPM at NO LOAD. But when you mount prop on it, RPM will be reduced. Here we will take example of two props 10x3.8 and 10x6. When you mount 10 inch diameter prop RPM of motor will be reduced to 3600 RPM "

This, in itself is not confusing, but what I'm struggling with, is where the calculated drop came from (3600RPM) for a 10" prop (6660 / 1.85 = 3600).

Is there another calculation that needs to be carried out for say, a 9" or 8" prop to get the divisible factor?

Many Thanks in advance

Mark

 

[Charles Bruckner]

Mark.
There are a lot of other factors that can affect performance.
Air temp.
humidity
altitude.

I am new to the hobby but I am not new to engineering.
In this calculation there should also be a viable for pitch of the prop in addition to its diameter.

 

[Viper1953]

Hi Charles,

It makes perfect sense to include air temp, humidity, and altitude. However, I'm trying to follow the calculations in the article but for different size props.

The next part of the calc in the article includes the pitch.

Heres the calculation breakdown for a 10in prop:

No load RPM / 1.85 = RPM with prop​

RPM with Prop / 60 = Revs per sec with prop​

Revs per sec with prop * prop pitch = climb rate (inch/sec)
​

The figure highlighted in red (1.85) is not stated in the article, but it is the divisible factor to get from 6660 to 3600 RPM (for a 10in prop).

It is this that I am struggling to understand - where did that figure come from?

Cheers

Mark

 

[Charles Bruckner]

Thanks for the better explanation,

With out knowing what that variable is the equation is almost useless.